Assignment PhysicsII 02
description
Transcript of Assignment PhysicsII 02
Department of PhysicsIndian Institute of Technology Kharagpur
Kharagpur-721302, West Bengal, India
Subject No. PH20003(PHYSICS II)
Assignment Due date : 14th
August 2015 Total Marks: 100
Thursday 6th August, 2015
Assignment # 2
§1. Answer the following questions(a) A hydrogen atom is 0.53 A in radius. Estimate the minimum energy an electron can have in thisatom. [use Heisenberg uncertainty principle](b) Estimate the lowest energy of a neutron confined in one-dimensional infinitely high potential boxof length 10−14 m.
§2. Consider the wave function Ψ(x) = C√x e−(x/2a) , where ‘C’ is normalization constant, describes a
particle in the range (0, ∞). Find the expectation value for x and x2, and most probable value of xwhere probability density is maximum.
§3. A particle is located in a 1-D square potential well of width ‘L’ with absolutely impenetrable walls(0 < x < L).Find the probability of staying within a region L/3 ≤ x ≤ 2L/3, for the particle (a) in the groundstate, (b) in the first excited state.
§4. A particle of mass m moves in a potential well of width 2a. Its potential energy is infinite forx < −a & x > +a. Inside the region, −a < x < a, its potential energy is given byU(x) = −h2a2
m a2(a2−x2). In addition, the particle is in a stationary state that is described by the wave-
function Ψ(x) = A(1− x2/a2) for −a < x < +a and by Ψ(x) = 0 elsewhere.(a) Show that A = (15/16a)1/2
(b) Determine the energy of the particle in terms of h, m and a.(c) Find the probability that the particle is located between x = −a/3 & x = a/3.
§5. A photon with wavelength λ is absorbed by an electron confined to a box having infinite potentialwall. As a result, the electron moves from state n = 1 to n = 4.(a) Find the width of the box.(b) What is the wavelength of the photon emitted in the transition of that electron from the staten = 4 to the state n = 2?
§6. In a 1-D potential barrier (potential height V0, and length L) problem, A particle of mass m havingenergy E > V0 coming from −∞. Using boundary condition solve for R & T which will be a functionof length of the potential barrier.
R =(k21−k22)2 sin2 k2L
4k21k22+(k21−k22)2 sin2 k2L
where k1 =√
2mEh2 ; and k2 =
√2m(E−V0)
h2
T =4k21k
22
4k21k22+(k21−k22)2 sin2 k2L
1