Academic Session: Autumn 2015-16
Tutorial IA (Solutions 1a-A) –
Instruction – Not to be shared with students and the problems should
be tried in tutiorial class and homework
Answers to the Problems at the end of Part-I:
1. 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.
Ans: E3 – E2 = hv = hc/λ – 2.41 x 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
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2. What is the uncertainty in the position of electron, if uncertainty in its velocity is
0.0058 m/s?
Ans: x . v = h / 4πm
x = 6.02 x 10-34 / 4 x 3.14 x 9.1 x 10-34 x 0.0058
x = 0.01 m
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3. (a) What are the major flaws in Classical Mechanics?
(b) Give one example of experimental result which is at variance with classical
mechanics.
Ans: (a) The classical mechanical approach to the description of the behavior of a
system can be illustrated by two equations.
(i) Total energy of the particle in terms of its kinetic energy 1/2 mv 2 and potential
energy V(x) can be expressed as:
E = ½ mv2 + V(x), where x and V are function of time t.
p = mv (p = momentum)
Therefore, E = p2/2m + V(x) ………..Eq.1
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Equation one can be interpreted as follows:
If we know both momentum and position of the particle, we can calculate it’s total
energy.
p = mv = m(dx/dt). So, equation 1 is a differential equation for x as a function of time
t. If the particle is of fixed energy, it might be possible to solve the equation and one
can get the position & momentum as a function of time t. i. e. one can predict it’s
trajectory precisely.
(ii) The second basic equation is Newton’s second law, which is a relation between
the acceleration d2x/dt2 of a particle and the force F(x) it experiences.
We have from Newton’s second Law
p• = F(x), p• = m (dv/dt) = m (d2x/dt2)
That is, if we know the force acting on the particle in every region of space we might
be able to solve this equation and find it’s momentum at all the times, and from that
it’s position. Also, continuous variation of energy is possible.
However, certain experiments done in late 19 th century and early in this century gave
results, totally at variance with the predictions of classical physics. All however,
could be explained on the basis that, classical physics is wrong in allowing systems to
possess arbitrary amounts of energy.
(b) photoelectric effect
In photoelectric effect, when a metal surface is irradiated with light, electrons are
emitted. It was shown that, (i) no electrons are ejected, if the frequency of the
incident light does not exceeds a threshold value, that is characteristic of each metal,
regardless of the intensity of the light. (ii) Kinetic energy of ejected electron is
linearly proportional to the frequency (figure 1).
K. E.
(iii) Even at low intensities of light, electrons are ejected immediately if the
frequency is above the threshold frequency.
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Academic Session: Autumn 2015-16
(c) Atomic spectra : The atomic spectra of hydrogen was proved to be line spectra, it
is further clarify that electrons energy states are discontinuous and hence, de-
excitation provides line spectra.
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4. Work function of sodium is 2.5 eV. Predict whether the wavelength 6500Å is
suitable for a photoelectron ejection or not. Guess and conclude the event.
Ans: Energy of incident light , hc/λ= 6.62 x 10-34 x 5 x 108\6500x 10-10
= 3.055 x 10-19 J
= 1.9 eV
Conclusion: The energy is lower than the work function, hence, no ejection will
occur.
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5. A particle of mass 3 10-20 kg is moving with a velocity 2 107 m sec-1. Find out
the wavelength associated with the particle.
Ans:
= h/p = h/mv or = 6.626 10-34 J s /3 10-20 kg 2 107 m sec-1
or = 1.1043 10-20 m sec-1 (Units: J = Nm; N = kg m s-2)
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6. Write down the Hamiltonian for the following: H2+.
Ans:
H = -h2/82m 2 – e2/ra - e2/rb + e2/Rab
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7. Calculate the energy per photon for radiation of wavelength (a) 600 nm, (b) 550
nm, and (c) 400 nm.
Ans: (a) E = h = hc/ = (6.626 10-34 Js 2.998 108 m s-1 /600 10-9 m )
6.02 1023 = 199.3 kJ
E = h = hc/ = (6.626 10-34 Js 2.998 108 m s-1 /550 10-9 m ) 6.02 1023 =
183.67 kJ
E = h = hc/ = (6.626 10-34 Js 2.998 108 m s-1 /400 10-9 m ) 6.02 1023 =
133.58 kJ
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Academic Session: Autumn 2015-16
8. Calculate the de Broglie wavelength of a He atom traveling at 1000 m s-1.
Ans: = h/p = h/mv
= 6.626 10-34 Js / 4 1.66054 10-27 kg 1000 m s-1
or = 9.9756 10-27 m s-1
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9. The work function of certain metal is 3.44 10-18 J. The absorption of a photon of
unknown wavelength ionizes ejects electrons with velocity 1.03 106 m s-1. What is
the wavelength of the incident radiation?
Ans: KE of the electron = ½ 9.10939 10-31 kg (1.03 106 ms-1)2 = 4.8321 10-
19
J
4.8321 10-19 J = h -
4.8321 10-19 J = (6.626 10-34 Js ) - 3.44 10-18 J
= (4.8321 10-19 J + 34.4 10-19 J)/ 6.626 10-34 Js
= 5.920 10-15 s-1
= c/ = 2.998 108 m s-1/ 5.920 10-15 s-1 = 50.6 nm
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10. What is the de-Broglie wavelength of electron having K.E. of 5 eV?
Ans:
K.E. = 1/2 mv2
v =√ 2K.E / m
Now = λ = h / mv
= h / √2K.E.m
= 6.62 x 10-34 / √ 2 x 5 x 1.6 x 10-19 x 9.1 x 10-31
= 5.486 x 10-10 m
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11. A diode laser used in fiber-optic telecommunication technologies (E.g.
Broadband internet) has a wavelength of λ TC = 1310 nm and a typical output
power of PTC = 5 milliwatt (mW). How many photons does such a laser put out per
second?
Answer: 3.28X1016 photons/s.
Calculation:
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12. Find the photon energy in electron volts for an electromagnetic wave of frequency
(a) 100 MHz in the FM radio band and (b) 900 kHz in the AM radio band.
For f = 100 MHz: E = hf = (6.626 10-34 Js)(100 106 s-1)
= 6.626 10-26 J 1eV/1.602 10-19J
= 4.14 10-7 eV
(b) For f = 900 kHz E = hf = (6.626 10-34 Js)(900 103 s-1)
= 5.963 10-28 J 1eV/1.602 10-19J
= 3.72 10-9 eV
13.“He2 cannot exist in ground state, but it can exists in excited state” – Explain.
Ans: The electronic configuration of He2 is 1s21s*2, so the bond order is zero,
thus it cannot exists. But when one electron is excited from 1s* level to 2s the
bond order becomes one, so in the excited state, He2 can exists. (Draw the MO
diagram and show it).
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14.
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