Q.83. A particle of mass m is located in a threedimensional cubic potential well with absolutely impenetrable walls. The side of the cube is equal to a. Find:
(a) the proper values of energy of the particle;
(b) the energy difference between the third and fourth levels;
(c) the energy of the sixth level and the number of states (the degree of degeneracy) corresponding to that level.
Ans. We proceed axactly as in (6.81). The wave function is chosen in the form
= A sin k_{1}x sin k_{2 }y sin k_{3} z .
(The origin is at one comer of the box and the axes of coordinates are along the edges.) The boundary conditions are that for
x = 0, x = a, y = 0, y = a, z = 0, z = a
This gives
The energy eigenvalues are
The first level is (1, 1, 1). The second has (1, 1, 2), (1, 2, 1) & (2, 1, 1). The third level is (1, 2, 2) or (2, 1, 2) or (2, 2, 1). Its eneigy is
The fourth energy level is (1, 1, 3) or (1, 3, 1) or (3, 1, 1)
Its eneigy is
(b) Thus
(c) The fifth level is (2, 2, 2). The sixth level is (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)
Its eneigy is
and its degree of degeneracy is 6 (six).
Q.84. Using the Schrodinger equation, demonstrate that at the point where the potential energy U (x) of a particle has a finite discontinuity, the wave function remains smooth, i.e. its first derivative with respect to the coordinate is continuous.
Ans. We can for definiteness assume that the discontinuity occurs at the point x = 0. Now the schrodinger equation is
We integrate this equation around x = 0 i.e., from where are small positive numbers. Then
or
Since the potential and the energy E are finite and ψ(x) is bounded by assumption, the integral on the right exists and
Thus
So is continuous at x = 0 (the point where U (x) has a finite jump discontinuity.)
Q.85. A particle of mass m is located in a unidimensional potential field U (x) whose shape is shown in Fig. 6.2, where U(0) = . Find:
(a) the equation defining the possible values of energy of the particle in the region E < U_{0}; reduce that equation to the form
where Solving this equation by graphical means, demonstrate that the possible values of energy of the particle form a discontinuous spectrum;
(b) the minimum value of the quantity l^{2}U_{0} at which the first energy level appears in the region E < U_{0}. At what minimum value of l^{2}U_{o }does the nth level appear?
Ans.
(a) Starting from the Schrodinger equation in the regions l & II
(1)
(2)
where U_{0} > E > 0 , we easily derive the solutions in I & II
(3)
(4)
where
The boundary conditions are
(5)
andare continuous at x = /, and ψ must vanish at x = + .
Then
and
so
From this we get
or
(6)
Plotting the left and right sides of this equation we can find the points at which the straight lines cross the sine curve. The roots of the equation corresponding to the eigen values of energy E_{i }and found from the inter section points (kl)_{i }, for which tan (kl)_{i} < 0 (i.e. 2^{nd }& 4^{th} and other even quadrants). It ii seen that bound states do not always exist. For the first bound state to appear (refer to the line (b) above)
(b) Substituting, we get
as the condition for the appearance of the first bound state. The second bound state will appear when Id is in the fourth quadrant The magnitude of the slope of the straight line must then be less than
Corresponding to
For n bound states, it is easy to convince one self that the slope of the appropriate straight line (upper or lower) must be less than
Then
Do not forget to note that for large n both + and  signs in the Eq. (6) contribute to solutions.
Q.86. Making use of the solution of the foregoing problem, determine the probability of the particle with energy E = U_{0}/2 to be located in the region
Ans.
and
or
It is easy to check that the condition of the boud state is satisfied. Also
Then from the previous problem
By normalization
The probability of the particle to be located in the region x > l is
Q.87. Find the possible values of energy of a particle of mass m located in a spherically symmetrical potential well U (r) = 0 for r < r_{o } and U (r) = for r = r_{0}, in the case when the motion of the particle is described by a wave function ψ(r) depending only on r.
Instruction. When solving the Schrodinger equation, make the substitution ψ(r) = x (r)/r.
Ans. The Schrodinger equation is
when ψ depends on r only,
If we put
and Thus we get
The solution is
and
(For r < r_{0} we have rejected a term 5 cos k r as it does not vanish at r = 0). Continuity of the wavefunction at r = r_{0} requires
Hence
Q.88. From the conditions of the foregoing problem find:
(a) normalized eigenfunctions of the particle in the states for which ψ(r) depends only on r;
(b) the most probable value r_{pr }for the ground state of the particle and the probability of the particle to be in the region r < r_{pr}.
Ans. (a) The nomalized wave functions are obtained from the normalization
Hence
(b) The radial probability distribution function is
For the ground state n = 1
so
By inspection this is maximum for . Thus
The probability for the particle to be found in the region r < r_{pr} is clearly 50 % as one can immediately see from a graph of sin^{2}x.
Q.89. A particle of mass m is located in a spherically symmetrical potential well U (r) = 0 for r < r_{0} and U (r) = U_{0} for r >
(a) By means of the substitution ψ(r) = x(r)/r find the equation defining the proper values of energy E of the particle for E < U_{0}, when its motion is described by a wave function ψ(r) depending only on r. Reduce that equation to the form
(b) Calculate the value of the quantity at which the first level appears.
Ans. If we put
the equation for X(r) has the from
which can be written as
and
where
The boundary condition is
These are exactly same as in the one dimensional problem in problem (6.85) Wc therefore omit further details
Q.90. The wave function of a particle of mass in in a unidimensional potential field U (x) = kx^{2}/2 has in the ground state the form ψ(x) = Ae^{αx2}, where A is a normalization factor and a is a positive constant. Making use of the Schrodinger equation, find the constant a and the energy E of the particle in this state.
Ans. The Schrodinger equation is
We are given
Then
Substituting we find that following equation must hold
since the bracket must vanish identicall. This means that the coefficient of x^{2} as well the term independent of x must vanish. We get
Putting k/m = ω^{2}, this leads to
Q.91. Find the energy of an electron of a hydrogen atom in a stationary state for which the wave function takes the form ψ(r) = A (1 + ar) e^{αr}, where A, a, and α are constants.
Ans. The Schrodinger equation for the problem in Gaussian units
In MKS units we should read
we put (1)
We are given that
so
Equating the coefficients of r^{2}, r, and constant term to zero we get
(2)
(3)
(4)
From (3) either a = 0,
In the first case
This state is the ground state.
It n the second, case
This state is one with n = 2 (2s).
Q.92. The wave function of an electron of a hydrogen atom in the ground state takes the form ψ(r) = Ae^{r}/r_{1}, where A is a certain constant, r_{1} is the first Bohr radius. Find: (a) the most probable distance between the electron and the nucleus;
(b) the mean value of modulus of the Coulomb force acting on the electron;
(c) the mean value of the potential energy of the electron in the field of the nucleus.
Ans. We first find A by normalization
since the integral has the value 2.
Thus
(a) The most probable distance r_{pr} is that value of r for which
is maximum. This requires
or
(b) The coulomb force being given by , the mean value o f its modulus is
In MKS units we should read
(c)
In MKS units we should read
Q.93. Find the mean electrostatic potential produced by an electron in the centre of a hydrogen atom if the electron is in the ground state for which the wave function is ψ(r) = Ae^{r}/r_{1}, where A is a certain constant, r_{1} is the first Bohr radius.
Ans. We find A by normalization as above. We get
Then the electronic charge density is
The potential due to this charge density is
so at the origin
Q.94. Particles of mass m and energy E move from the left to the potential barrier shown in Fig. 6.3. Find:
(a) the reflection coefficient R of the barrier for E > U_{0};
(b) the effective penetration depth of the particles into the region x > 0 for E < U_{o}, i.e. the distance from the barrier boundary to the point at which the probability of finding a particle decreases efold.
Ans. (a) We start from the Schrodinger equation
which we write as
and
It is convenient to look for solutions in the form
In region I (x < 0), the amplitude of is written as unity by convention. In II we expect only a transmitted wave to the right, B = 0 then. So
The boundary conditions follow from the continuity of
1 + R = A
Then
The reflection coefficient is the absolute square of R :
(b) In this case is unchanged in form but
we must have B = 0 since otherwise ψ(x) will become unbounded as Finally
Inside the barrier, the particle then has a probability density equal to
This decreases to of its value in
Q.95. Employing Eq. (6.2e), find the probability D of an electron with energy E tunnelling through a potential barrier of width l and height U_{0} provided the barrier is shaped as shown:
(a) in Fig. 6.4;
(b) in Fig. 6.5.
Ans. The formula is
Here V(x_{2}) = V(x_{1}) = E and V(x) >E in the region x_{2} > x > x_{1}.
(a) For the problem, the integral is trivial
(b) We can without loss of generality take x = 0 at the point the potential begins to climb. Then
Then
Q.96. Using Eq. (6.2e), find the probability D of a particle of mass m and energy E tunnelling through the potential barrier shown in Fig. 6.6, where U(x) = U_{0}(1 — x^{2}/l^{2}).
Ans. The potential is The turning points are
Then
The integral is
Thus
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