Atomic Structure Solved Exercise

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Solved text book exercise chapter 5 atomic structure chemistry book 1 for FSC pre medical and pre engineering and for All board of intermediate and secondary education of Pakistan.
    TEXT BOOK EXERCISE
Q1.      Select the most suitable answer for the given one.
i.          The nature of the positive rays depends on
            (a)        The nature of the electrode
            (b)        The nature of the discharge tube
            (c)        The nature of the residual gas
            (d)       all of the above
ii.         The velocity of photon is
            (a)        Independent of its wavelength
            (b)        Depends on its wavelength
            (c)        Equal to square of its amplitude
            (d)       Depends on its source
iii.        The wave number of the light emitted by a certain source is 2 x 106 m. the wavelength of this light will be
(a)        500nm                         (b)        500m
(c)        200nm                         (d)       5 x 107 m
iv.        Rutherford’s model of an atom failed because
(a)        The atom did not have a nucleus an electron
(b)        It did not account for the attraction between protons and neutrons
(c)        It did not account for the stablility of the atom
(d)       There is actually no space between the nucleus and the electrons
v.         Bohr model of atom is contradicted by
(a)        Planck quantum theory
(b)        Pauli’s exclusion principle
(c)        Heisenberg’s uncertainty principle
(d)       all of the above
vi.        Splitting of spectral lines when atoms are subjected to strong electric field is called.
(a)        Zeeman effect                         (b)        Stark effect
(c)        Photoelectric effect     (d)       Compton Effect
vii.       In the ground state of an atom, the electron is present
(a)        In the nucleus              (b)        in the second shell
(c)        Nearest to the nucleus
(d)       farthest from the nucleus
viii.            Quantum number values for 2p orbitals are
(a)        n=2, l=l                       (b)        n=1, l=2
(c)        n=1, l=0                       (d)       n=2, l=0
ix.        Orbitals having same energy are called
            (a)        Hybrid orbitals            (b)        valence orbitals
            (c)        Degenerate orbitals     (d)       d-orbitals
x.         when 6d orbitals is complete, the entering electron goes into

            (a)        7f         (b)        7s         (c)        7p        (d)       7d
Ans.    (i)c                   (ii)a                  (iii)a                 (iv)c                 (v)c
            (vi)b                 (vii)c                (viii)a               (ix)c                 (x)c
Q.2      Fill in the blanks with suitable words
(i)         B-particles are nothing but­­­ _______moving with a very high speed.
(ii)        The charge on one mole of electrons is ________coulombs.
(iii)             The mass of hydrogen atom is _________grams.
(iv)             The mass of one mole of electron is _________.
(v)               Energy is ________when electron jumps from higher to a lower orbit.
(vi)             The ionization energy of hydrogen atom can be given by formula ________.
(vii)           For d sub-shell, the azimuthal quantum number has a value of ________.
(viii)         The number of electrons in a given sub-shell is given by formula  ________.
(ix)             The electronic configuration of H- is ________.

Ans.    i)electrons                    ii)96500                       iii)1066x10-24               iv)5.484x10-7kg             v)emitted                     vi) Bohr                       vii)2                             viii)2(2l+1)                  ix) 1s2
Q.3      Indicate true or false as the case may be.
(i)         A neutron is slightly lighter particle than a proton.
(ii)        A photon is the massless bundle of energy but has momentum.
(iii)             The unit of Rydberg constant is the reciprocal of unit of length.
(iv)             The actual isotopic mass is a whole number.
(v)               Heisenberg’s uncertainty principle is applicable to macroscopic bodies.
(vi)             The nodal plane in an orbital is the plane of zero electron density.
(vii)           The number of orbitals present in a sublevel is given by the formula (2l_1)
(viii)         Magnetic quantum number was introduced to explain Zeeman and stark effects.
(ix)             Spin quantum number tells us the direction of spin of electron around the nucleus.
Ans.    i)False                          ii)True              iii)True                         iv)False                        v)False             vi)True                         vii)True                        viii)True           ix)Ture
Q.4      Keeping in mind the discharge tube experiment , answer the following questions.
(a)        Why is it necessary to decrease the pressure in the discharge tube to get the cathode rays?
Ans.    There will be no flow of current through the gas when the pressure in the discharge tube is high. In the presence of high pressure, the cathode rays will not flow from the cathode surface.
(b)       Which ever gas is used in the discharge tube the nature of the cathode rays remains the same. Why?
Ans.    Cathode rays are composed of negatively charged particles (electrons). They are constituents of all gases. So, cathode rays are independent of the nature of the gas in the discharge tube.
(c)        Why e/m value of the cathode rays is equal to that of electron?
Ans.    Cathode rays are composed of electrons, so their e/m value is just equal to that of electron.
(d)       How the bending of the cathode rays in the electric and magnetic fields shows that they are negatively charged?
Ans.    when cathode rays are passed through an electric field created by two charged metal plates, they are deflected towards the positively charged plate. This shows that they are negatively charged.
                        When cathode rays are passed between the poles of a magnet, the magnet neither attracts nor repels but cause them to move in a curved path perpendicular to the line drawn between the poles of the magnet. This shows that they are negatively charged.
(e)        Why positive rays are also called canal rays?
Ans.     Positive rays pass through canals or holes in the cathode, so they are called canal rays.
(f)        The e/m value of positive rays for different gases are different but those for cathode rays the e/m values is the same Justify it.
Ans.    The e/m value for positive rays are different for different gases because they differ in mass. the mass of the positive particles is the same as that of the atom or molecule form which it is created. Heavier the gas, smaller the e/m value.
                        The e/m value for cathode rays is the same because cathode rays are composed of electrons which have constant mass.
(g)               The e/m value for positive rays obtained from hydrogen gas is 1836 time less than that of cathode rays. Justify it.        
Ans.    The e/m value for positive rays obtained form hydrogen gas is 1836 time less than that of cathode rays. This is because the mass of proton which is created from H-atom is 1836 time more than that of an electron (cathode rays particle).
Q.5      (a)        Explain Milliken’s oil drop experiment to determine the charge of an electron.
(b)        What is J.J Thomson’s experiment for determining e/m value of electron?
            (c)        Evaluate mass of electron from the above two experiments.
Q.6      (a)        Discuss Chadwick’s experiment for the discovery of neutrons. Compare     the properties of electron, proton, and neutron.
(b)        Rutherford’s atomic model is based on the scattering of a-particles from a thin gold foil. Discuss it and explain the conclusions.
Q.7      (a)        Give the postulates of Bohr’s atomic model. Which postulate tell us that     orbits are stationary and energy is quantized?
(b)        Derive the equation for the radius of nth orbit of hydrogen atom using Bohr’s model.
(c)        How does the above equation tell you that?
(i)         Radius is directly proportional to the square of the number of orbit.
(ii)        Radius is inversely proportional to the number of proton in the nucleus.
Ans.    The equation for the radius of nth orbit is:
                                                r= 
Since, , h, ,m and e are constant , therefore , the factoris constant.
Therefore,                                r=constant x
   Or
                                                r = 
            Hence, we can say:
(i)         Radius is directly proportional to the square of the number of orbit.
(ii)        Radius is inversely proportional to the number of protons in the nucleus
(d)       How do you come to know that the velocities of electrons in higher orbits are less than those in lower orbits of hydrogen atom?
Ans.     According to Bohr, since the electron keeps on revolving around the nucleus:
            Therefore, centrifugal force=Electrostatic force of attraction
                                                =
or                                             r=
For H-atom,Z=1
                                            r=
 Since, e,,and m are constant, therefore, the factor is constant.
Therefore,                                r=constant x 
                                                r=
            Hence, the radius of a moving electron is inversely proportional to the square of its velocity. The smaller the radius of the orbit, the higher is the velocity of electron. Hence, the velocities of electrons in higher orbits are less than those in lower orbits of hydrogen atom.
(e)        Justify that the distance gaps between different orbits go on increasing form the lower.
Ans.     We know that:                        r=0.529 x[n2]
            When n=1                   r1=0.529
            When n=2                   r2=0.529x4=2.11
            When n=3                   r3=0.529x9=4.75
            When n=4                   r4=0.529x16=8.4
            When n=5                   r5=0.529x25=13.22
            Distance between orbits are:
                                    r2 –r1=(2.11 – 0.529) =1.581
                                     r3 –r2=(4.75 – 2.11) =2.64
                                                r4 –r3=(8.4 – 4.75) =3.65
                                                r5 –r4=(13.22 – 8.4) =4.82
            From the data of radius difference, it is clear that the distance gaps between different orbits go on increasing from the lower to the higher orbits.
Q8.      Derive the formula for calculating the energy of an electron in nth orbit using Bohr’s model. Keeping in view this formula explain the following:
(a)        The potential energy of the bounded electron is negative.
Ans.     According to Bohr, the energy of electron is calculated from the equation:
                                    En=- 2.178 x 10-18[]
                        When               n=,               then En=0
            Consider that an electron is present at an infinite distance from the nucleus, so there is no interaction between the two. The energy of this electron is zero.
            Now, suppose that the electron moves closer and closer to the nucleus. Since electron is negatively charged and nucleus is positively charged, no work needs to be done on the electron. The electron can move towards the nucleus by itself due to electrical attraction. Thus, work is done by the electron itself as it moves towards the nucleus. As a result, the potential energy falls, i.e. it become less than zero. Any value less than zero is negative. Hence, the potential energy of electron becomes more and more negative as the electron moves closer and closer towards the nucleus.
(b)       Total energy of the bounded electron is negative.
Ans.    When the electron is at infinite distance from the nucleus, there is no electrostatic interaction between the two. Therefore, the energy of the system in this state is assumed to be zero. As the electron moves closer to the nucleus, it does some work and energy is of the electron becomes negative. The negative value of energy would keep increasing as the electron moves to the energy levels nearer to the nucleus. the negative value of total energy shows that electron is bound by the nucleus i.e, electron is under the force of attraction of the nucleus.
(c)        Energy of an electron is inversely proportional to n2 , but energy of higher orbits are always greater than those of the lower orbits.
Ans.    Energy of electron:
            The energy of electron in different orbits can be calculated by using the following equation:
                                    En = - kJmol-1 
            Energy of an electron is inversely proportional to n2.
            When n =1      E1=              =1312.36kJ mol-1
                When n =2      E2=              =328.09kJ mol-1
                When n =3      E3=              =145.82kJ mol-1
                When n =4      E4=              =82.023kJ mol-1
                When n =5      E1=              =52.49kJ mol-1
            The energy of an electron is inversely proportional to n2 .
            As the value of ‘n’ increases, the value of energy increases. The energy of higher orbits are always greater than those of the lower orbits.
                        E5 >E4 >E3> E2 >E1
(d)       The energy difference between adjacent levels goes on decreasing sharply.
Ans.    The energy difference between adjacent levels can be found as:
            E=E2 – E1 =(- 328.09) – (-1312.36) =984.27kJmol-1
                E=E3 – E2 =(- 145.82) – (- 328.09) =182.27kJmol-1 
            E=E4 – E3 =(- 82.023) – (- 145.82) =63.797kJmol-1
                E=E5 – E4 =(- 52.49) – (- 82.023) =29.533kJmol-1
            From the data of energy difference, it is found that the energy difference between adjacent levels goes on decreasing sharply.
Q9.      (a)        Derive the following equation for hydrogen atom which are related to:
i.                    Energy difference between two levels, n1 and n2 .
ii.                  Frequency of photon emitted when an electron jumps from n2 to n1 .
iii.                Wave number of the photon when the electron jumps from n2 to n1 .
(b)        Justify that Bohr’s equation for the wave number can explain the spectral lines of Lyman, Blamer and paschen series. 
Q10.    (a)        What is spectrum? Differentiate between continuous spectrum, and line spectrum.
(b)        Comparison between line emission and line absorption spectra.
Ans.   
Line emission spectrum
Line absorption spectrum
1.       “An atomic spectrum which consists of bright lines against a dark background is called line emission spectrum. “
2.    it is produced when radiations emitted by an excited substance are analysed in a spectroscope.
1.       “An atomic spectrum which consists of bright lines against a dark background is called line emission spectrum. “
2.        it is produced when white is passed through the gaseous element and the transmitted rays are analysed in a spectroscope.
            (c)        What is the origin of the line spectrum?
Q11.    (a)        Hydrogen atom and He+ are monoelectronic system, but the size of He+ is much smaller than H+ , why?
Ans.    H-atom and He+ are monoelectronic system. It means both H-atom and He+ have one electron in the valence shell. H-atom has one proton in the nucleus whereas He+ has two proton in the nucleus. So, the force of attraction between two protons and one electron is greater than one proton and one electron. Hence, the size of He+ is much smaller than H-atom.

               Also, we know that:    r=0.529 
               The size of H-atom:    r=0.529=0.2645   
               The size of He+ ion:     r=0.529=0.529 x =0.2645  
Hence, the size of He+ is much smaller than H-atom. This is because the nucleus of He+ has greater attraction for the electron as compared to H-atom which contains one proton in the nucleus.
(b)   Do you think that the size of li-2+ is even smaller than HE+ ? Justify with calculation.
Ans.       The size of He+ ion:              r=0.529=0.2645 
                  The size of li-2+ ion:              r=0.529=0.1763         
                  The size of li-2+ ion is much smaller than the size of He+ ion:
Q12.    (a)        What are X-rays? What is their origin? How was the idea of atomic number derived from the discovery of X-rays?
(b)               How does the Bohr’s model Justify the Moseley‘s equation?
Q13.    Point out the defects of Bohr’s model. How these defects are partially covered by dual nature of electron and Heisenberg’s uncertainty principle?
Q14.    (a)        Briefly discuss the wave mechanical model of atom. How has it given the idea of orbital. Compare orbit and orbital.
Ans.    Comparison between orbit and orbital:
Orbit
Orbital
1.       “A circular path around the nucleus in which the electron revolves is called an orbit.”
2. It is circular in Shape.
3.  It represents that an electron moves around the nucleus in one plane, i.e., in a flat surface.
4.   It is against Heisenberg’s uncertainty principle.
5.  The maximum number of electrons in an orbit is 2n2, where ‘n’ is the number of the orbit.
1.       “The volume of space within an atom in which there is 95% chance of finding an electron is called orbital.”
2.  It may be spherical, dumbbell or double dumbbell in shape.
3.   It represent that an electron can move around the nucleus in three dimensional space.
4.  It is in accordance with Heisenberg’s uncertainty principle.
5. The maximum number of electrons in an orbital is two.
            (b)        What are quantum number? Discuss their significance.
            (c)        When azimuthal quantum number has a value 3, then there are seven value of magnetic quantum number. Give reasons.
Q15.    Discuss rules for the distribution of electrons in energy sub-shells and in orbitals. (a)          What is (n+ l ) rule. Arrange the orbitals according to this rule. Do you think that this rule is applicable to degenerate orbitals?
(b)               Distribute electrons in orbitals of 57La, 29Cu, 79Au, 24Cr, 53I, 86Rn.
Ans.       Electronic configurations:
57La=1s2 2s2 2p6 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f4 (Actual configuration)
=1s2 2s2 2p6 3p6 4s2 3d10 4p6 5s2 4d10 5s2 5p6 5d1(Expected configuration)
29Cu =1s2 2s2 2p6 3d9 4s2                                                (Actual configuration)
=1s2 2s2 2p6 3p6 4s2 3d10 4s1                               (Expected configuration)
79Au=1s2 2s2 2p6 3p6 4s2 3d10 4p6 5s2 4d10 4f145s2 5p6 5d10 gs1
24Cr=1s2 2s2 2p6 3s2 3p6 3d4 4s2                                                (Actual configuration)
=1s2 2s2 2p6 3s2 3p6 3d4 4s1                                        (Expected configuration)
53I=1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p5
 86Rn==1s2 2s2 2p6 3s2 3p6 3d4 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6
From the above configuration, it is important to note that there are three irregularities in the general trend. The electronic configuration of Cr and Cu show deviation from the expected configuration.
Expected Configurations:                    Cr=1s2 2s2 2p6 3s2 3p6 3d4 4s1
                                                                                Cu=1s2 2s2 2p6 3s2 3p6 3d4 4s1
Actual Configurations:                        Cr=1s2 2s2 2p6 3s2 3p6 3d4 4s2
                                                                                Cu=1s2 2s2 2p6 3s2 3p6 3d4 4s2
                    This is because the half filled and fully filled configurations (i.e.,d5,d10,f7,f14 )have lower energy ot more stability. Thus, in order to become more stable, one of 4s electrons goes into 3d orbitals so that 3d orbitals get half filled or fully filled configuration in Cr and Cu respectively.
Reasons for stability of Half filled and fully filled orbitals:
1.            Exchange energy; All the orbitals in a given sub-shell  have equal energies. The electrons present in different orbitals of the same sub-shell can exchange their positions. Such an exchange of electrons results in release of energy called exchange energy . Greater the exchange energy, more is the stability associated with the orbitals. It has been observed that in case of exactly half-filled and fully filled orbitals the electrons can exchange their positions more readily as compared to other configurations. As a result, half-filled and fully filled configurations are more stable.
2.            Symmetry: If removal or addition of an electron results in the symmetrical distribution of electrons in an orbital, the electronic configuration becomes more stable. Therefore, half-filled and fully filled configurations in which each orbital contains one and two electrons respectively , are more stable.
Irregular Structure of La: The electronic configuration of La is irregular and does no t follow the general trend. This is because strong nuclear attraction and less shielding effect caused by d and f electrons. In this case remember before adding any electron in the 4f orbital, a single electron is added to a 5d orbital. The remaining nine electrons enter the 5d sub-shell after the 4f sub-shell has been completely filled with fourteen electrons. Similarly, one electron enters the 6d sub-shell before any electron enters the 5f sub-shell.
Q16.    Draw the shapes of s, p and d-orbitals. Justify these by keeping in view the azimuthal and magnetic quantum numbers.
Q17.    A photon of light with energy 10-19 j is emitted by a source of light/
            (a)        Convert this energy into the wavelength, frequency and wave number of the photon in terms of meters, hertz and m-1 respectively.
Solution:
                        E=10-19 J                      h=6.625 x 10-34 js
            Formula used:             E=hv
                                    V=
                                     V=
                                    V=1.51 x 1014 s-1
                                    V=1.51 x 1014 Hz                  [CPS=s-1 =Hz]
            Now,               =
                                    E=10-19 J=10-19 kg m2 s-2  [1 J=1 kf m2 s-2 ]
                                    h =6.625 x 10-34 kg m2 s-2 s
                                       =6.625 x 10-34 kg m2s-1
                                                = 
                                    =19.875x 10-7 m
                                    =1.9875 x 10-6 m=1.9875 x 10-4 cm
            Now,                           
                                                
                                                =0.51 x 106 m-1
                                                =5.1 x 105 m-1 =5.1 x 103 cm-1
(b)        Convert this energy of the photon into ergs and calculate the wave length in cm, frequency in Hz and wave number in cm-1 .
            h=6.625x 10-34 js or 6.625 x 10-27 ergs. C=3 x 108 ms-1 or 3x 10 +10 cms-1 .
Solution:
            E= 10-19 J =10-19 x 107 erg [1 j =107 erg]
                                    =10-12 erg
            =1.98 x 10-6 m=1.98 x10-6 x 102 cm=1.98 x 10-4 cm
            =5x105 m-1 =5 x 105 x 10-2 cm-1 =5x 103 cm-1
Q18.    The formula for calculating the energy of an electron in hydrogen atom gives by Bohr, s model
                                    En =
            Calculate the energy of the electron in first orbit of hydrogen atom. The values of various parameters are same as provided in Q.19.
Solution:
                        n =                               E=?
            Formula:          En =-2.178 x 10-18 []J
                                    En=-2.178  x 10-18 []J
                                    E=-2.178 x 10-18 J
                                    E=-2.18 x 10-18 J
Q19.    Bohr’s equation for the radius of nth orbit of electron in hydrogen atom is
                                                rn =
(a)        When the electron moves from n=1 to n=2, how much does the radius of the orbit increases.
Solution:
               =8.85 x10-12 C2 J-1 m-1        ;           m=9.108 x 10-31 kg
               h=6.624 x 10-34 js               ;           e=1.602 x 10-19 C        ;=3.14
Form       n=1 to n=2   :           J=kg m2 s-2     :           C=    s-1
Formula: rn =
            r1= 
               r1=
            r1=
               r1= 5.29 x 10-11 jm-1 s2 kg-1
                    r1= 5.29 x 10-11 kg m2 s2m-1 s2  kg-1
                    r1= 5.29 x 10-11 m 
               r1= 5.29 x 10-1
               r1= 0.529 
Also          r2= 0.529 [22]
               R2= 0.529 x 4 =2.116
Increase in radius, (r2  - r1)       =2.116- 0.529
                                                                =1.587
(b)         What is the distance traveled  by th electron when it goes from n=2 to n =3and n =9 to n=10?
               =8.85x 10-12 C2 J-1 m-1 , h=6.24x10-34 js,  =3.14,
               m=9.108 x10-31 kg , C=1.602 x 10-19 C
               While doing calculations take care of units of energy parameter.
               J=kgm2 s-2 , c=kg m  s-1
Solution:                     r=0.529 (n2)
               For    n=2      r2 =0.529(22) =2.116 
                        n =3     r3=0.529(32)  =4.761      
               Distance traveled by the electron when it goes from n=9 to n =3
                                    r3 – r2 =4.761  - 2.116 =Answer
For          n=9   r9 =0529(92)=42.849
               n =10            r10=0.529 (102)=52.9
Distance traveled by the electron when it goes grom n=9 to n=10
                          r10 – r9 =52.9 - 42.849 =10.051  Answer
Q20.       Answer the following questions, by performing the calculation s.
(a)    Calculate the energy of first five orbits of hydrogen atom an determine the energy differences between them.
(b)    Justify that energy difference between second and third orbits is approximately five times smaller than that between first and second orbits.
(c)    Calculate the energy of electron in He+ in first five orbits and justify that the energy differences are different from those of hydrogen atom.
(d)   Do you think that group of the spectral lines of He+ are at different places than those for hydrogen atom? Give reasons.
Q21.    Calculate the value of principal quantum number if an electron in hydrogen atom revolves in an orbit of energy - 0.242 x 10-18 j.
Solution:
                                    E= - 242 x 10-18 j    ;                n=?
            Formula:          E=- 2.178 x 10-18 
                                    n2= - 2.178 x 10-18 x 
                                    n2=- 2.178 x 10-18x
                                    n2=
                                    n2=9
                                    n= 3 Answer
Q22.    Bohr’s formula for the energy levels of hydrogen atom for any system say H, He+, Li-2+ , etc is
                                    En=
            Or
                                    En= - k []
            For hydrogen Z=1 and for He+ , Z=2
(a)          Draw an energy level diagram for hydrogen atom and He+ .
(b)         Thinking that k = 2.18 x 10-18 j, calculate the energy needed to remove the electron from hydrogen atom and from He+ .
Solution:
               For H;           Z=1 ; n=1 ; k=2.18 x 10-18 j
               Formula        En =-k[]
                                    E1 = -2.18 x 10-18 []
                                    E1=-2.18 x 10-18 j
                                    =-2.18 x 10-18 [] J
                                    =-2.18 x 10-18 0j=0
               Now energy required to remove an electron from the orbit:
                                                = - E1
                                                =0 – (-8.72 x10-18  J)
                                                =8.72 x 10-18 J Answer
(c)        How do you justify that the energies calculated in (b) are the ionization energies of H and He+ ?
Ans.    The energy difference between first and infinite levels of energy for H atom is  2.18x 10-21 kJ and for He+ ion is 8.72x10-21 kJ are the ionization energies of H and He+ respectively. These values are the same as determined experimentally.
(c)                Use Avogadro’s number to convert ionization energy values in kJ mol-1 or H and He+ .
Solution:
               For H:                       =2.18 x 10-18 J
   The value of energy obtained for the electron is in J/atom. If this quantity is multiplied by Avogadro’s number and divided by 1000, the value of E will become.
                                    = 
                                                =13.1236 x 102 kJ mol-1
                                                =1312.36 kJ mol-1Answer                                                          For He+ ;          =8.72 x 10-18 J/atom
                                    =8.72 x 10-18 x
                                                =52.4944 x 102 kJ mol-1
                                                =5249.44 kJ mol-1 Answer         
(e)        The experimental values of ionization energy of H and He+ are 1331 kJ mol-1 and 5250 kJ mol-1respectively. How do you compare your values with experimental values ?
Q23.    Calculate the wave number of the photon when the electron jumps from
            i.          n=5 to n=2.
            ii.         n=5 to n=1.]
            In which series of spectral lines and spectral regions these photons will appear.
Solution:
(i)         n=5 to n =1 .   =?
Formula :                     =1.09678 x 107 [-]m-1
                                    =1.09678x 107[-]m-1
                                    =1.09678x 107[]m-1
                                    =1.09678x 107[]m-1
                                                =1.09678x 107m-1     =2.3x106 m-1 Answer
            The photon will appear in the Balmer series
ii)                  n =5 to n= 1
Solution:
               n = 5 to n=1                         ;                       =?
               Formula:                   =1.09678 x 107 [-]m-1
                                                                =1.09678 x 107 [-]m-1
                                                                =1.09678 x 107 [-]m-1
                                                                =1.09678 x 107 []m-1
                                                                =1.09678 x 107 m-1 Answer
               The photon will appear in the Lyman series.
Q24.    A photon of a wave number 102.70 x 105 m-1 is emitted when electron jumps form higher to n=1.
(a)        Determine the number of that orbit from where the electron falls.
Solution:
                                    =102.70 x 105              ;           n2=?
Formula used:             =1.09678 x 107 [-]m-1
                                102.70x 105 m-1 =1.09678 x 107 [-]m-1
                                =[-]
                                    0.936=1-   
                                    =1-0.936
                                    =0.064
                                    =
                                    =16
                                    n2 =          =4 Answer
(b)        Indicate the name of the series to which this photon belongs.
Ans.    Since the electron falls from n=4 to n=1, therefore, the name of the series is Lyman series
(d)               If the electron will fall from higher orbit to n=2, then calculate the wave number of the photon emitted. Why this energy difference is so small as compared to that in part (a)?
Solution:
                                    n1 =2                ;           n2=4
                                    =?
            Formula:          =1.09678 x 107 [-]m-1
                                                =1.09678 x 107 [-]m-1
                                                =1.09678 x 107 [-]m-1
                                                =1.09678 x 107 []m-1
                                                =0.20565x 107 m-1
                                                =0.20565x 105m-1 Answer
Q25.    (a)        What is de Broglie’s wavelength of an electron in meters traveling at half a speed of light ?
                                       m =9.109 x 10-31 kg , c=3x 108 ms-1
Solution:
            M=9.109 x 10-31 kg   :           v=x3x 108 ms-1         =1.5x 108 ms-1
                    h =6.624 x 10-34 js    =6.624 x 10-34 kg m2 s-1
            Formula:          =
                        =
                        =0.485 x 10-11 m
                        =0.485 x 10-12 m
                        =0.485 x 10-12 x 1010 
                        =0.485 x 10-2       =0.0485  Answer
(b)        Convert the mass of electron into grams and velocity of light into cms-1 , and then calculate the wavelength of an electron in cm.
Solution:
            m=9.109 x 10-31 kg =9.109 x 10-31 x 103 g =9.109 x 10-28 g
            c =3 x 10-8 ms-1 =3 x 10-8 x 102 cms-1 =3 x 108 cms-1
            =4.85 x 10-12 m=4.85 x 10-12 x102 cm =4.85 x 10-10 cm
                                                                        =0.048 x 10-8 cm Answer
(c)                Convert the wavelength of electron from meters to
               i)       nm       ii)                 iii)        pm

Solution:
i) =4.85 x 10-12 m=4.85 x 10-12 x109 cm =4.85 x 10-3 nm=0.048nm
ii) =4.85 x 10-12 m=4.85 x 10-12 x1010=4.85 x 10-2 =0.0485  Ans.
iii)        =4.85 x 10-12 m=4.85 x 10-12 x10-12  pm =4.485 x 10-24 pm  Ans.

I am Mechatronics and Control Engineer from UET, Lahore. Helping Engineering students gives me pleasure.

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