Chapter 27, Problem 74P

(a)

To determine

The stopping potential of incident light.

Answer to Problem 74P

The stopping potential of incident light is $1.2\text{V}$.

Explanation of Solution

Write the expression to find the maximum kinetic energy.

  $K_{\max }=e{V}_s$                                                                                                             (I)

Here, $K_{\max }$ is the maximum kinetic energy, $V_s$ is the stopping potential.

Write the expression for Einstein’s photoelectric equation.

  $K_{\max }=\frac{hc}{\lambda }-\varphi$                                                                                                       (II)

Here, $h$ is the plank’s constant, $c$ is the speed of light in vacuum, $\lambda$ is the wavelength of light, $\varphi$ is the work function.

Equate equations (I) and (II) to find the work function for sodium.

  $e{V}_s=\frac{hc}{\lambda }-\varphi$                                                                                                       (III)

Re-arrange the expression to find the work function for sodium.

  $\varphi =\frac{hc}{\lambda }-e{V}_s$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, $570\text{nm}$ for $\lambda$ and $0.28\text{V}$ for $V_s$ to find the work function for sodium.

  $\begin{array}{c}\varphi =\frac{1240\text{eV}\cdot \text{nm}}{570\text{nm}}-e\left(0.28\text{V}\right)\\ =2.18\text{eV}-0.28\text{eV}\\ =1.9\text{eV}\end{array}$

Re-arrange the equation (III) to find the stopping potential.

  $V_s=\frac{hc}{e\lambda }-\frac{\varphi }{e}$

Conclusion:

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, $400\text{nm}$ for $\lambda$ and $1.9\text{eV}$ for $\varphi$ to find the stopping potential.

  $\begin{array}{c}{V}_s=\frac{1240\text{eV}\cdot \text{nm}}{e\left(400\text{nm}\right)}-\frac{1.9\text{eV}}{e}\\ =3.1\text{V}-1.9\text{V}\\ \text{=}1.2\text{V}\end{array}$

(b)

To determine

The stopping potential for the incident light when the intensity is $2.0{\text{W/m}}^{\text{2}}$.

Answer to Problem 74P

The stopping potential for the incident light when the intensity $2.0{\text{W/m}}^{\text{2}}$ is $0.28\text{V}$.

Explanation of Solution

Write the equation to find the stopping potential.

  $V_s=\frac{hc}{e\lambda }-\frac{\varphi }{e}$

The stopping potential of incident light is independent on intensity of light as in the above expression. Thus, the stopping potential of light cannot be changed when the intensity of the light is increased from $1.0{\text{W/m}}^{\text{2}}$ to $2.\text{0}{\text{W/m}}^{\text{2}}$.

Conclusion:

The stopping potential of the incident light is $0.28\text{V}$.

(c)

To determine

Work function if the sodium.

Answer to Problem 74P

Work function if the sodium is $1.9\text{eV}$.

Explanation of Solution

Write the expression to find the maximum kinetic energy.

  $K_{\max }=e{V}_s$                                                                                                            (I)

Here, $K_{\max }$ is the maximum kinetic energy, $V_s$ is the stopping potential.

Write the expression for Einstein’s photoelectric equation.

  $K_{\max }=\frac{hc}{\lambda }-\varphi$                                                                                                       (II)

Here, $h$ is the plank’s constant, $c$ is the speed of light, $\lambda$ is the wavelength of light, $\varphi$ is the work function.

Equate equations (I) and (II) to find the work function for sodium.

  $e{V}_s=\frac{hc}{\lambda }-\varphi$                                                                                                       (III)

Re-arrange the expression to find the work function for sodium.

  $\varphi =\frac{hc}{\lambda }-e{V}_s$

Conclusion:

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, $570\text{nm}$ for $\lambda$ and $0.28\text{V}$ for $V_s$ to find the work function for sodium.

  $\begin{array}{c}\varphi =\frac{1240\text{eV}\cdot \text{nm}}{570\text{nm}}-e\left(0.28\text{V}\right)\\ =2.18\text{eV}-0.28\text{eV}\\ =1.9\text{eV}\end{array}$