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Found in: Page 1215

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 39-29 a shows a thin tube in which a finite potential trap has been set up where ${{\mathbf{V}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{}}{\mathbf{V}}$. An electron is shown travelling rightward toward the trap, in a region with a voltage of ${{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{9}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{V}}$, where it has a kinetic energy of 2.00 eV. When the electron enters the trap region, it can become trapped if it gets rid of enough energy by emitting a photon. The energy levels of the electron within the trap are ${{\mathbf{E}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{,}}{{\mathbf{E}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}$, and ${{\mathbf{E}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{eV}}$, and the non quantized region begins at ${{\mathbf{E}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{9}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{eV}}$as shown in the energylevel diagram of Fig. 39-29b. What is the smallest energy such a photon can have?

The energy of the photon is 7eV.

See the step by step solution

## Step 1: Introduction:

An electron is shown travelling rightward toward the trap, in a region with a voltage ${{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{9}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{V}}$, where it has a kinetic energy of ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{eV}}$.

## Step 2: Determine the energy of the photon:

The electron losses some energy when it jumps from the quantized region to the non-quantized region. The energy of the photon is the sum of the kinetic energy and the potential energy, the potential energy is equal to the difference between the third and fourth levels.

The kinetic energy is 2 eV therefore, the energy of the photon is,

$\mathrm{E}=\mathrm{K}+\mathrm{PE}\phantom{\rule{0ex}{0ex}}=\mathrm{K}+∆\mathrm{E}$

Since, the change in energy is equal to ${\mathrm{E}}_{4}-{\mathrm{E}}_{3}$.

Substitute ${\mathrm{E}}_{4}-{\mathrm{E}}_{3}$ for $∆\mathrm{E}$ in the above equation.

$\mathrm{E}=\mathrm{K}+\left({\mathrm{E}}_{4}-{\mathrm{E}}_{3}\right)$

Substitute 2 eV for K, $9\mathrm{eV}$ for role="math" localid="1661767190577" ${\mathrm{E}}_{4}$ and 4 eV for ${\mathrm{E}}_{3}$ in the above equation.

$E=\left(2eV\right)+\left(9eV\right)-\left(4eV\right)\phantom{\rule{0ex}{0ex}}=7\mathrm{eV}$

Hence, the energy of the photon is 7 eV.