Chapter 7: Q43E (page 281)

Explicitly verify that the simple function $R (r) = A e^{b r}$ can be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.

Short Answer

Expert verified

The angular momentum is zero.

The energy is $- \frac{m e^{4}}{8 π r_{0} h^{2}}$

Step by step solution

Given data

$R (r) = A^{e^{- b r}}$

Here, A and b constants.

Concept

The property of any rotating body, which is described as the product of the moment of inertia and the angular velocity of rotation of the body, is known as angular momentum. It is analogous to the linear momentum for rotatory motion.

The radial equation is,

$- \frac{h^{2}}{2 m} \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r}) R (r) + \frac{h^{2} I (I + 1)}{2 m r^{2}} R (r) - \frac{1}{4 π ε_{0}} \frac{e^{2}}{r} R (r) = E R (r) . . . . . . (1)$ (1)

Where,

m is the mass

e is the magnitude of the charge

I is the angular momentum

$ε_{0}$ is the permittivity of free space.

To determine angular momentum and energy

Substitute $A e^{- b r}$ for $R (r)$ in equation (1).

$- \frac{h^{2}}{2 m} \frac{1}{r^{2}} (r^{2} \frac{d}{d r}) A e^{- b r} + \frac{h^{2} I (I + 1)}{2 m r^{2}} A e^{- b r} - \frac{1}{4 π ε_{0}} \frac{e^{2}}{r} A e^{b r} = E A e^{b r}$

Now, solve the localid="1659322395860" $\frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r}) A e^{- b r}$ term as follows:

localid="1659322573731" $\frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r}) A e^{- b r} = \frac{1}{r^{2}} \frac{d}{d r} (r^{2} (- b)) A e^{- b r} = \frac{- b}{r^{2}} (2 r) + \frac{- b}{r^{2}} (r^{2}) (- b A e^{- b r}) = \frac{- b}{r^{2}} (A e^{- b r}) (2 r) + b^{2} A e^{- b r} = - b A e^{- b r} (\frac{1}{r} (2 - b r))$

Substitute $- b A e^{- b r} (\frac{1}{r} (2 - b r))$ for $\frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r}) A e^{- b r}$ in the equation (1).

$- \frac{h^{2}}{2 m} (- b A e^{- b r} (\frac{1}{r} (2 - b r))) + \frac{h^{2} I (I + 1)}{2 m r^{2}} A e^{- b r} - \frac{e^{2}}{r} A e^{b r} = E A e^{b r} (\begin{matrix} - \frac{h^{2} (- b A e^{- b r})}{2 m} (\frac{2}{r}) (- \frac{h^{2} (- b A e^{- b r})}{2 m}) (- b) + \frac{h^{2} I (I + 1)}{2 m r^{2}} A e^{- b r} - \frac{1}{4 π ε_{0}} \frac{e^{2}}{r} A e^{- b r} \\ = E A e^{b r} \end{matrix}) - \frac{h^{2} b (A e^{- b r})}{m r} - \frac{h^{2} (A e^{- b r}) b^{2}}{2 m} + \frac{h^{2} I (I + 1)}{2 m r^{2}} A e^{- b r} - \frac{1}{4 π ε_{0}} \frac{e^{2}}{r} A e^{- b r} = E A e^{- b r} . . . . . . . . . . . . . (2)$

The last three terms in the above equation are proportional to the terms in the radial equation given by equation (1).

Comparing the coefficients of the terms in the above equation with equation (1), we get-

$\frac{h^{2} I (I + 1)}{2 m r^{2}} A e^{- b r} = \frac{h^{2} I (I + 1)}{2 m r^{2}} A e^{- b r}$

On solving we get,

I =0

Therefore, the angular momentum is zero.

The potential energy term in equation (2) is equal to $- \frac{h^{2} (A e^{- b r}) b}{2 m r}$

The potential energy term in equation (1) is equal to $- \frac{1}{4 π ε_{0}} \frac{e^{2}}{r} (A e^{- b r})$

Thus equate the above two expressions and solve as follows:

$\frac{h^{2} (A e^{- b r}) b}{m r} = - \frac{1}{4 π ε_{0}} \frac{e^{2}}{r} (A e^{- b r}) \frac{h^{2} b}{m} = \frac{e^{2}}{4 π ε_{0}} b = \frac{m e^{2}}{4 π ε_{0} h^{2}}$

Now, equate the potential energy term to total energy, as the kinetic term is zero.

$- \frac{2 h^{2} (A e^{- b r}) b}{m r} = E A e^{b r} - \frac{h^{2} (A e^{- b r}) b}{m r} = E A e^{b r} E = - \frac{h^{2} b^{2}}{2 m}$

Substitute $\frac{m e^{2}}{4 π ε_{0} h^{2}}$ for b.

role="math" localid="1659324115735" $E = h^{2} r {(\frac{m e^{2}}{4 π ε_{0} h^{2}})}^{2} = - \frac{m e^{4}}{8 π r_{0} h^{2}}$

Therefore, the angular momentum is zero and the energy is $- \frac{m e^{4}}{8 π r_{0} h^{2}}$

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Explicitly verify that the simple function $R (r) = A e^{b r}$ can be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.

Short Answer

Step by step solution

Given data

Concept

To determine angular momentum and energy

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Explicitly verify that the simple function Rr=Aebrcan be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.

Short Answer

Step by step solution

Given data

Concept

To determine angular momentum and energy

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physics of Motion

Further Mechanics and Thermal Physics

Electromagnetism

Quantum Physics

Geometrical and Physical Optics

Thermodynamics

Study anywhere. Anytime. Across all devices.

Explicitly verify that the simple function $R (r) = A e^{b r}$ can be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.