Chapter 27, Problem 41P

To determine

(a) To Determine:

The momentum of an electron.

Answer to Problem 41P

Solution:

Momentum of the electron is $1.47\times 10\text{kg}\text{m/s}$.

Explanation of Solution

Given:

The de-Broglie wavelength of an electron: $4.5\times {10}^{-10}\text{m}$

Formula Used:

de-Broglie Wavelength;

$\lambda =\frac{h}{p}$

Calculation:

By using the formula $\lambda =\frac{h}{p}$

The momentum of an electron can be calculated as

$\begin{array}{l}\lambda =\frac{h}{p}\\ p=\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{4.5\times {10}^{-10}\text{m}}=1.47\times 10\text{kg}\text{m/s}\end{array}$

To determine

(b) To Determine:

The speed of an electron.

Answer to Problem 41P

Solution:

Speed of the electron is $1.61\times {10}^6\text{m/s}$.

Explanation of Solution

Given:

The de-Broglie wavelength of an electron: $4.5\times {10}^{-10}\text{m}$

Formula Used:

$\lambda =\frac{h}{p}=\frac{h}{mv}$

Calculation:

By using the formula $\lambda =\frac{h}{mv}$

The speed of an electron can be calculated as

$\begin{array}{l}v=\frac{h}{m\lambda }\\ \lambda =\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{(9.11\times {10}^{-31}\text{kg})(4.5\times {10}^{-10}\text{m})}=1.61\times {10}^6\text{m/s}\end{array}$

To determine

(c) To Determine:

Voltage needed to accelerate the electron from rest to $1.61\times {10}^6\text{m/s}$ speed.

Answer to Problem 41P

Solution:

Voltage needed to accelerate the electron from rest to $1.61\times {10}^6\text{m/s}$ speed is 6 V.

Explanation of Solution

Given:

The de-Broglie wavelength of an electron: $4.5\times {10}^{-10}\text{m}$

Speed of the electron:$1.61\times {10}^6\text{m/s}$.

Formula Used:

$\text{KE}=\frac{1}{2}m{v}^2$

Calculation:

By using the formula $\text{KE}=\frac{1}{2}m{v}^2$

The kinetic energy of an electron can be calculated as

$\begin{array}{l}\text{KE}=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}\left(9.11\times {10}^{-31}\text{kg}\right){(}^{1.61}=11.80\times {10}^{-19}\text{J}\end{array}$

Now on converting the joules to electron volts, this is equal to $\frac{11.80\times {10}^{-19}\text{J}}{1.06\times {10}^{-19}\text{J/eV}}=6.0\text{eV}$

When an electron is accelerated by 6 V potential differences. It gains the energy of $6.0\text{eV}$ which is needed to accelerate the electron from rest to $1.61\times {10}^6\text{m/s}$ speed.