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### Fundamentals Of Physics

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

# An electron is confined to a narrow-evacuated tube of length 3.0 m; the tube functions as a one-dimensional infinite potential well. (a) What is the energy difference between the electron’s ground state and its first excited state? (b) At what quantum number n would the energy difference between adjacent energy levels be 1.0 ev-which is measurable, unlike the result of (a)? At that quantum number, (c) What multiple of the electron’s rest energy would give the electron’s total energy and (d) would the electron be relativistic?

(a) $∆E=1.3×{10}^{-19}eV$

(b) $n=1.2×{10}^{19}$

(c) $\frac{E}{{E}_{rest}}=1.2×{10}^{13}$

(d) Yes

See the step by step solution

## Step 1: Identification of the given data

The given data is listed below as-

The length of the narrow-evacuated tube L = 3m

## Step 2: The energy equation is given by

The energies difference is given by the equation-

${\mathbf{∆}}{\mathbit{E}}{\mathbf{=}}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}\left({n}_{2}^{2}-{n}_{1}^{2}\right)$

Here, L is the length of the tube.

## Step 3: To determine the energy difference between the electron’s ground state and its first excited state  (a)

The difference between energies is given by the equation:

$∆E=\frac{{h}^{2}}{8m{L}^{2}}\left({n}_{2}^{2}-{n}_{1}^{2}\right)$

For, ${n}_{2}=2$ and ${n}_{1}=1$

$∆E=\frac{\left(6.626×{10}^{-34}\right)}{8×9×109×{10}^{-31}×{3}^{2}}\left({2}^{2}-{1}^{2}\right)\phantom{\rule{0ex}{0ex}}=2×{10}^{-38}J\phantom{\rule{0ex}{0ex}}=1.3×{10}^{-19}eV$

Thus, the energy difference between the electron’s ground state and its first excited state is $=1.3×{10}^{-19}eV$ .

## Step 4: At what quantum number n would the energy difference between adjacent energy levels be 1.0 ev-which is measurable (b)

The difference between energies is given by equation:

$∆E=\frac{{h}^{2}}{8m{L}^{2}}\left[{\left(n+1\right)}^{2}-{n}^{2}\right]$

The value of n is obtained by solving the above equation as below:

$n=\frac{4∆m{L}^{2}}{{h}^{2}}-\frac{1}{2}$

Substituting all the values in above equation:

$n=\frac{4×1.602×{10}^{-19}×9.109×{10}^{-31}×{3}^{2}}{\left(6.626×{10}^{-34}\right)}-\frac{1}{2}\phantom{\rule{0ex}{0ex}}n=1.2×{10}^{19}$

Thus, at quantum number $n=1.2×{10}^{19}$ the energy difference between adjacent energy levels will be 1.0 ev-which is measurable.

## Step 5: To determine the multiple of the electron’s rest energy that would give the electron’s total energy (c)

At $n=1.2×{10}^{19}$ , the energy equation is given as

$∆E=\left(\frac{{h}^{2}}{8m{L}^{2}}\right){n}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{\left(6.626×{10}^{-34}\right)}^{2}}{8×9.109×{10}^{-31}×{3}^{2}}\left(1.2×{10}^{19}\right)\phantom{\rule{0ex}{0ex}}=0.96J$

The rest energy is given as:

${E}_{rest}=m{c}^{2}\phantom{\rule{0ex}{0ex}}=9.109×{10}^{-31}×3×{10}^{8}\phantom{\rule{0ex}{0ex}}=8.2×{10}^{-14}J$

Now, to find multiple of the electron’s rest energy that would give the electron’s total energy, divide the value of E by ${E}_{rest}$ .

$\frac{E}{{E}_{rest}}=\frac{1.13}{8.2×{10}^{-14}}\phantom{\rule{0ex}{0ex}}=1.2×{10}^{13}$

## Step 6: To determine whether the electron is relativistic (d)

Yes, the electron is relativistic because the kinetic energy of this electron would give the nonrelativistic equation which suggests that its speed is greater than the speed of light.