FIGURE is the velocity-versus-time graph for a 2.0 kg

Chapter 11, Problem 13E

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

One measure of the intensity of a transition of frequency v is the oscillator strength, f, which is defined as

\(f=\frac{8 \pi^2 m_{\mathrm{e}} v\left|\boldsymbol{\mu}_{\mathrm{f}}\right|^2}{3 h e^2}\)

Consider an electron in an atom to be oscillating harmonically in one dimension (the three-dimensional version of this model was used in early attempts to describe atomic structure). The wavefunctions for such an electron are those in Table 9.1. Show that the oscillator strength for the transition of this electron from its ground state is exactly \(\frac{1}{3}\).

Questions & Answers

QUESTION:

One measure of the intensity of a transition of frequency v is the oscillator strength, f, which is defined as

\(f=\frac{8 \pi^2 m_{\mathrm{e}} v\left|\boldsymbol{\mu}_{\mathrm{f}}\right|^2}{3 h e^2}\)

Consider an electron in an atom to be oscillating harmonically in one dimension (the three-dimensional version of this model was used in early attempts to describe atomic structure). The wavefunctions for such an electron are those in Table 9.1. Show that the oscillator strength for the transition of this electron from its ground state is exactly \(\frac{1}{3}\).

ANSWER:

Step 1 of 5

We have to determine the work done on the object during each of the four intervals AB, BC, CD, and DE.

The work done on the object during each of the four intervals AB, BC, CD, and DE can be determined by WorkEnergy Theorem.

WorkEnergy Theorem states that “the total work done an object is equal to change in the kinetic energy of the body”.

\(W =\Delta K E\)

\(=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\)

Where,

\(m =\text { mass of the object }\)

\(=2 \mathrm{~kg}\)

\(v_{i} =\text { initial velocity in } \mathrm{m} / \mathrm{s}\)

\(v_{f} =\text { final velocity } \mathrm{in} \mathrm{m} / \mathrm{s}\)

 

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back