Chapter 27, Problem 42P

To determine

(a) To Determine:

The wavelength of an electron of energy $10\text{eV}$.

Answer to Problem 42P

Solution:

The wavelength of an electron of energy $10\text{eV}$ is $0.39\text{nm}$.

Explanation of Solution

Given:

Energy of an electron is: $10\text{eV}$

Formula Used:

$\begin{array}{l}KE=\frac{p^2}{2m}\\ \lambda =\frac{h}{p}\end{array}$

Calculation:

By using the formula $KE=\frac{p^2}{2m}$, the value of p can be derived as $p={\left[2m(KE)\right]}^{\frac{1}{2}}$

Now using the formula for the wavelength as $\lambda =\frac{h}{p}$, the wavelength of the electron can be calculated as $\lambda =\frac{h}{p}$

$\begin{array}{c}\lambda =\frac{h}{p}=\frac{h}{{\left[2m(KE)\right]}^{\frac{1}{2}}}=\frac{hc}{{\left[2m{c}^2(KE)\right]}^{\frac{1}{2}}}\\ =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{{\left[2\left(0.511\times {10}^6\text{eV}\right)(10\text{eV})\right]}^{\frac{1}{2}}}=0.39\text{nm}\end{array}$

$0.39\text{nm}$ is the wavelength of the electron with energy $10\text{eV}$

To determine

(b) To Determine:

The wavelength of an electron of energy $100\text{eV}$.

Answer to Problem 42P

Solution:

The wavelength of an electron of energy $100\text{eV}$ is $0.12\text{nm}$.

Explanation of Solution

Given:

Energy of an electron is: $100\text{eV}$

Formula Used:

$\begin{array}{l}KE=\frac{p^2}{2m}\\ \lambda =\frac{h}{p}\end{array}$

Calculation:

By using the formula $KE=\frac{p^2}{2m}$, the value of p can be derived as $p={\left[2m(KE)\right]}^{\frac{1}{2}}$

Now using the formula for the wavelength as $\lambda =\frac{h}{p}$, the wavelength of the electron can be calculated as $\lambda =\frac{h}{p}$

$\begin{array}{c}\lambda =\frac{h}{p}=\frac{h}{{\left[2m(KE)\right]}^{\frac{1}{2}}}=\frac{hc}{{\left[2m{c}^2(KE)\right]}^{\frac{1}{2}}}\\ =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{{\left[2\left(0.511\times {10}^6\text{eV}\right)(100\text{eV})\right]}^{\frac{1}{2}}}=0.12\text{nm}\end{array}$

$0.12\text{nm}$ is the wavelength of the electron with energy $100\text{eV}$

To determine

(c) To Determine:

The wavelength of an electron of energy $1.0\text{keV}$

Answer to Problem 42P

Solution:

The wavelength of an electron of energy $1.0\text{keV}$ is $0.039\text{nm}$.

Explanation of Solution

Given:

Energy of an electron is: $1.0\text{keV}$

Formula Used:

$\begin{array}{l}KE=\frac{p^2}{2m}\\ \lambda =\frac{h}{p}\end{array}$

Calculation:

By using the formula $KE=\frac{p^2}{2m}$, the value of p can be derived as $p={\left[2m(KE)\right]}^{\frac{1}{2}}$

Now using the formula for the wavelength as $\lambda =\frac{h}{p}$, the wavelength of the electron can be calculated as $\lambda =\frac{h}{p}$

$\begin{array}{c}\lambda =\frac{h}{p}=\frac{h}{{\left[2m(KE)\right]}^{\frac{1}{2}}}=\frac{hc}{{\left[2m{c}^2(KE)\right]}^{\frac{1}{2}}}\\ =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{{\left[2\left(0.511\times {10}^6\text{eV}\right)(1.0\times {10}^3\text{eV})\right]}^{\frac{1}{2}}}=0.039\text{nm}\end{array}$

$0.039\text{nm}$ Is the wavelength of the electron with energy $1.0\text{keV}$