Solved Questions of Atomic Structure Part I
Why Bohr’s orbits are called stationary states?
This is because the energies of orbits in which the electrons revolve are fixed.
Explain why the electronic configuration of Cu is 3d104s1 and not 3d94s2.
In the 3d104s1 the d-sub shell is completely filled which is more stable.
Fe3+ ion is more stable than Fe2+ ion. Why?
In Fe3+ ion 3d sub shell is half filled hence more stable configuration.
Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength of 0.005 nm.
v = h/mλ
ev = 1/2mv2
Putting the values we get
V = 32.85 volt
Give one example of isodiapheres.
Isodiapheres have same difference between the number of neutrons and protons. For example
3919K & 3115P
Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n = 6 to n = 4 transition in He+. [Neglect reduced mass effect].
v-He+ = RZ2 [ 1/42 - 1/62]
= 4R [ 36 - 16/36 x 16 ] = 5R/36
v-H = R x 12 [ 1/22 - 1/n2]
∴ v-He+ = v-H
5R/36 = R/4 - R/n2
On solving above equation
n2 = 9
∴n = 3
Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as that of 6 → 4 transition in He+.
Calculate ionization potential in volts of (a) He+ and (b) Li2+
I.E. = 13.6Z2/n2
= 13.6 x Z2 \12 [Z =2 for He+]
= 13.6 x 4 = 54.4 eV
Similarly for Li2+ = 13.6 x 32/12
= 13.6 x 9 = 122.4 eV
Calculate the ratio of K.E and P.E of an electron in an orbit?
K.E. = Ze2/2r
P.E. = -Ze2/r
∴ P.E. = –2K.E
∴ K.E/P.E = - 1/2
How many spectral lines are emitted by atomic hydrogen excited to
nth energy level?
Thus the number of lines emitted from nth energy level
= 1 + 2 + 3 +………… n – 1 = ∑(n – 1)
∑n = n(n+1)/2
∴ ∑ (n – 1) = ( n-1) (n-1+1)/2 = (n-1) (n)/2
Number of spectral lines that appear in hydrogen spectrum when an electron jumps from nth energy level = n (n-1)/2
Solved Examples of Atomic Structure Part II
Calculate (a) the de Broglie wavelength of an electron moving with a velocity of 5.0x 105 ms–1 and (b) relative de Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity (h = 6.63 x 10–34 kg m2 s–1)
(a) λ = h/mv = 6.63 x 10-34 kgm2s-2/ (9.11 x 10-31kg) ( 5.0 x 105ms-1)
Wavelength λ = 1.46 x10–9m
(b) An atom of oxygen has approximately 16 times the mass of an atom of hydrogen. In the formula λ = h/mv, h is constant while the conditions of problem make v, also constant. This means that λ and m are variables and λ varies inversely with m. Therefore, λ for the hydrogen atom would be 16 times greater than λ for oxygen atom.
A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.
K.E. = P.E.
1/2mv2 = 1/4πεo Ze2/d
Hence d = Ze2/4πεo( 1/2mv2)
= 9 x 109 x 79 x ( 1.6 x 10-19)2 / ( 1.6 x 10-13 ) [ as 1 MeV = 1.6 x 10-13 J]
= 1.137 x 10-13 m
What is the wavelength associated with 150 eV electron
λ = h/√2 x m x K.E.
= 6.626 x 1-34Js / √ 2 x 9.1 x 10-31kg x 150 x 1.6 x 10-19 J = 6.626 x 10-34 / √4368 x 10-50 = 10–10 m = 1 Å
The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42 x 10–12 erg and –2.41 x 10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.
E3 – E2 = hv = hc/λ – 2.41 ´ 10–12 – (– 5.42 ´ 10–12) = 6.626 x 10-27 x 3 x 1010/ λ
∴ λ = 6.626 x 10-27 x 3 x 1010 / 3.01 x 10-12
= 6.604 x 10–5 cm = 6.604 x 10–5 x 108 = 6604 Å
O2 undergoes photochemical dissociation into one normal oxygen and one excited oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen atoms requires 498kJ mole–1. What is the maximum wavelength effective for photochemical dissociation of O2?
O2 —→ ON + Oexcited
O2 —→ ON + ON
E = 498 x 103 J / mole
= 498 x 103 / 6.023 x 1023J per molecule = 8.268 x 10–19 J
Energy required for excitation = 1.967 eV = 3.146 ´10–19J
Total energy required for photochemical dissociation ofO2
= 8.268 x 10–19 + 3.146 x 10–19 = 11.414 x 10–19 J hc/λ = 11.414 x 10–19J
λ = 6.626 x 10-34x 3 x 108 / 11.414 x 10-19= 1.7415 x 10–7 m = 1741.5 Å
Compare the wavelengths for the first three lines in the Balmer series with those which arise from similar transition in Be3+ ion. (Neglect reduced mass effect).
v-H = R x 12 ( 1/22 - 1/n2)
v-Be = R x 42 ( 1/22 - 1/n2)
∴ v-Be/v-H = λH / λBe = 16
So we can conclude that all transitions in Be3+ will occur at wavelengths 1/16 times the hydrogen wavelengths.