Chapter 27, Problem 47A

(a)

To determine

To Calculate: The maximum kinetic energy of the photoelectrons (in electron volts).

Answer to Problem 47A

Kinetic energy in electron volts is $5.0\text{ eV}$ .

Explanation of Solution

Given data:

  $V_0=5.0\text{ V}$

Formula used:

  $KE=-q{V}_0$

where q is the charge on an electron and $V_0$ is the stopping potential.

Calculation:

Kinetic energy is calculated as:

  $\begin{array}{l}KE=-q{V}_0\\ =-\left(-1\text{ e}\right)\left(5.0\text{ V}\right)\\ =5.0\text{ eV}\end{array}$

Conclusion:

Kinetic energy in electron volts is $5.0\text{ eV}$ .

(b)

To determine

To Calculate: The maximum kinetic energy of the photoelectrons (in joules).

Answer to Problem 47A

Kinetic energy in joules is $8.0\times {10}^{-19}\text{}\text{ J}$ .

Explanation of Solution

Given data:

  $V_0=5.0\text{ V}$

Calculation:

Since, $KE=5.0\text{ eV}$ (calculated in part a)

Kinetic energy in joules is calculated as:

  $\begin{array}{l}KE=\left(5.0\text{ eV}\right)\left(\frac{1.60\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)\\ =8.0\times {10}^{-19}\text{}\text{ J}\end{array}$

Conclusion:

Kinetic energy in joules is $8.0\times {10}^{-19}\text{}\text{ J}$ .