Q43E Explicitly verify that the simpl... [FREE SOLUTION]

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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 $- \frac{m e^{4}}{8 蟺 r_{0} h^{2}}$

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,

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

Where,

m is the mass

e is the magnitude of the charge

I is the angular momentum

$蔚_{0}$ is the permittivity of free space.

To determine angular momentum and energy

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

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

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).

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

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 $- \frac{1}{4 蟺蔚_{0}} \frac{e^{2}}{r} (A e^{- b r})$

Thus equate the above two expressions and solve as follows:

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

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 $\frac{m e^{2}}{4 蟺蔚_{0} h^{2}}$ for b.

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

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

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91影视

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

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

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

Astrophysics

Further Mechanics and Thermal Physics

Energy Physics

Quantum Physics

Kinematics Physics

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.