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Practice each question, then open the answer or worked solution.
Assuming an electron is confined to a 1nm wide region, find the uncertainty in momentum using the Heisenberg Uncertainty principle. You can assume the uncertainty in position ∆x as 1nm. Assuming p = ∆p, find the energy of the electron in electron volts.
As the electrons rotate in a circular path, ∆r = 1 nm = 10-9 m
∆p = h/∆x
∆p = (331/314) × 10-25
E = 1/2 mv2 = ∆p2/2m
E = 3.8 × 10-2 eV
Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?
Let nA be the no.of photons falling per second from beam A and nB be the no.of photons falling per second from beam B.
Given,
No.of photons hitting the screen by A is twice that of B
Therefore, nA = 2nB
Energy from the beam A = hvA
Energy from the beam B = hvB
Intensity = I = nAvA = nBvB
vB = 2vA
Two particles A and B of de Broglie wavelengths λ1 and λ2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).
According to de-Broglie wavelengths,
λ=h/p
p=h/ λ
p1 = h/ λ1
p2 = h/ λ2
p3 = h/ λ3
There are 4 possible cases and they are:
Case 1: When p1 and p2 are positive, then λ3 = λ1 λ1/ λ1+ λ2
Case 2: When p1 and p2 are negative, then λ3 = λ1 λ2/ λ1+ λ2
Case 3: When p1 > 0 and p2 < 0, then λ3 = λ1 λ2/ λ2 – λ1
Case 4: When p1 < 0 and p2 > 0, then λ3 = λ1 λ2/ λ1- λ2
A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1nm. The first maximum of intensity in the reflected beam occurs at θ = 30°. What is the kinetic energy E of the beam in eV?
From Bragg’s law, 2d sin θ = nλ
p = h/λ = 6.6 × 10-24 kg m/s
E = 1/2 mv2 = p2/2m
E = 0.085 eV
Long AnswersConsider a thin target (10–2m square, 10–3m thickness) of sodium, which produces a photocurrent of 100µA when a light of intensity 100W/m2 (λ = 660nm) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na = 0.97 kg/m3].
Area = 10-4 m2
Thickness = 10-3 m
Current = 10-4 A
Intensity = 100 W/m2
Mass = (volume)(density) = 0.97 × 10-4 gm
No.of target atoms = 0.254 × 1019
Total energy = nhv
Where n = 3.3 × 1016
The probability of emission of photoelectrons is 7.5 × 10-21
Consider an electron in front of the metallic surface at a distance d (treated as an infinite plane surface). Assume the force of attraction by the plate is given as 2 2 0 1 4 4 q πε d Calculate work in taking the charge to an infinite distance from the plate. Taking d = 0.1nm, find the work done in electron volts. [Such a force law is not valid for d < 0.1nm].
The figure shows that an electron is displaced slowly. The distance covered by the electron is x and this happens by the means of an external force
F = q2/4×4πε0d2
d = 0.1 nm = 10-10 m
Let x be the distance of the electron and the force of attraction on it be
Fx = q2/4×4πε0d2
Therefore, work done by the external force is integral of force from d to infinity.
Therefore, work done = 3.6 eV
A student performs an experiment on the photoelectric effect, using two materials A and B. A plot of Vstop vs ν is given in the figure.
(i) Which material A or B has a higher work function?
(ii) Given the electric charge of an electron = 1.6 × 10–19 C, find the value of h obtained from the experiment for both A and B. Comment on whether it is consistent with Einstein’s theory
i) The threshold frequency of A is
vOA = 5 × 1014 Hz
The threshold frequency of B is
vOB = 10 × 1014 Hz
Work function is ϕ = hv0
ϕOA/ ϕOB = 5 × 1014/10 × 1014 < 1
ϕOA < ϕOB
A particle A with a mass mA is moving with a velocity v and hits a particle B (mass mB) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.
According to the law of conservation of momentum,
mAv + mB0 = mAv1 + mBv2
mA(v-v1) = mBv2
1/2 mAv2 = 1/2 mAv12 + 1/2mBv22
mA(v-v1)(v+v1) = mBv22
v1 = (mA-mB/mA+mB)v
v2 = (2mA/mA+mB)v
λinitial = h/mAv
λfinal = h/mAv1
∆λ = λfinal – λinitial = h/mAv(mA+mB/mA-mB)-1
Consider a 20 W bulb emitting light of wavelength 5000 A ° and shining on a metal surface kept at a distance 2m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 A ° .
(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?
Given,
P = 20W
λ = 5000 Ao = 5000 × 10-10 m
d = 2m
ϕ0 = 2 eV
r = 1.5 × 10-10 m
No.of photon emitted per second by the bulb = n’ = dN/dt
i) No.of photon emitted by bulb per second = n’ = p λ/hc = 5 × 1019 sec
ii) Energy of the incident photon = hc/ λ = 2.48 eV
iii) Time required by the atomic disk to receive energy is = 28.4 sec
iv) No.of photons received by the atomic disk = 2
v) Photoelectric emission is instantaneous as the collision time between the free electron and the incident photon is very less.
The chapter Dual Nature of Radiation and Matter introduces students to the concept of matter that propagates through radiation. Under the close observation of a microscope, we can see that matter exists as both a wave, as well as a particle.
Some of the major topics covered in this chapter are