Chapter 17, Problem 11P

(a)

To determine

The speed of an electron with kinetic energy $750\text{eV}$ .

Answer to Problem 11P

The speed of the electron is $1.6\times {10}^7\text{m/s}$ .

Explanation of Solution

Given info:

Kinetic energy is $KE=750\text{eV}$ .

Formula used:

The kinetic energy is given by

  $KE=\frac{1}{2}m{v}^2$

Here, m is the mass and v is the speed.

Calculation:

The charge on electron is $e=-1.6\times {10}^{-19}\text{C}$ .

The mass of an electron is $m=9.11\times {10}^{-31}\text{kg}$ .

Substituting the given values

  $\begin{array}{c}750\times \left(-1.6\times {10}^{-19}\right)=\frac{1}{2}\left(9.11\times {10}^{-31}\right)v^2\\ v^2=\frac{2\times 750\times -1.6\times {10}^{-19}}{9.11\times {10}^{-31}}\\ v=1.6\times {10}^7\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the electron is $1.6\times {10}^7\text{m/s}$ .

(b)

To determine

The speed of an electron with kinetic energy $3.2\text{keV}$ .

Answer to Problem 11P

The speed of the electron is $3.35\times {10}^7\text{m/s}$ .

Explanation of Solution

Given info:

Kinetic energy is $KE=3.2\text{keV}$ .

Formula used:

The kinetic energy is given by,

  $KE=\frac{1}{2}m{v}^2$

Calculation:

The charge on electron is $e=-1.6\times {10}^{-19}\text{C}$ .

The mass of an electron is $m=9.11\times {10}^{-31}\text{kg}$

Substituting the given values

  $\begin{array}{c}\left(3.2\times {10}^3\right)\times \left(-1.6\times {10}^{-19}\right)=\frac{1}{2}\left(9.11\times {10}^{-31}\right)v^2\\ v^2=\frac{2\times 3.2\times {10}^3\times -1.6\times {10}^{-19}}{9.11\times {10}^{-31}}\\ v=3.35\times {10}^7\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the electron is $3.35\times {10}^7\text{m/s}$ .