Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Answer to Problem 39P

The value of the Planck’s constant is $6.93\times {10}^{-34}\text{J}\cdot \text{s}$ and the value of the work function of the metal is $6.37\times {10}^{-19}\text{J}$.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

  $K_{\max }=e{V}_0$        (I)

Here, $K_{\max }$ is the maximum kinetic energy of photon, $e$ is the charge on an electron and $V_0$ is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

  $E=\frac{hc}{\lambda }$        (II)

Here, $E$ is the energy of the photon, $h$ is the Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of photon.

Write the expression for the work function of the metal.

  $E=\varphi +K_{\max }$        (III)

Here, $\varphi$ is the work function of the metal.

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$ and $1.0\text{V}$ for $V_0$ in equation (I).

  $\begin{array}{c}{K}_{\max }=\left(1.6\times {10}^{-19}\text{C}\right)\cdot \left(1.0\text{V}\right)\\ =1.6\times {10}^{-19}\text{J}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $260\times {10}^{-9}\text{m}$ for $\lambda$ in equation (II).

  $\begin{array}{c}E=\frac{h\left(3\times {10}^8\text{m}/\text{s}\right)}{260\times {10}^{-9}\text{m}}\\ =\left(1.15\times {10}^{15}\right)h\text{J}\end{array}$

Substitute $\left(1.15\times {10}^{15}\right)h\text{J}$ for $E$ and $1.6\times {10}^{-19}\text{J}$ for $K_{\max }$ in equation (III).

  $\left(1.15\times {10}^{15}\right)h\text{J}=\varphi +\left(1.6\times {10}^{-19}\text{J}\right)$        (IV)

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$ and $2.3\text{V}$ for $V_0$ in equation (I).

  $\begin{array}{c}{K}_{\max }=\left(1.6\times {10}^{-19}\text{C}\right)\cdot \left(2.3\text{V}\right)\\ =3.68\times {10}^{-19}\text{J}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $207\times {10}^{-9}\text{m}$ for $\lambda$ in equation (II).

  $\begin{array}{c}E=\frac{h\left(3\times {10}^8\text{m}/\text{s}\right)}{207\times {10}^{-9}\text{m}}\\ =\left(1.45\times {10}^{15}\right)h\text{J}\end{array}$

Substitute $\left(1.45\times {10}^{15}\right)h\text{J}$ for $E$ and $3.68\times {10}^{-19}\text{J}$ for $K_{\max }$ in equation (III).

  $\left(1.45\times {10}^{15}\right)h\text{J}=\varphi +\left(3.68\times {10}^{-19}\text{J}\right)$        (V)

Subtract the equation (IV) from (V).

  $\begin{array}{c}0.3h\times {10}^{15}=2.08\times {10}^{-19}\\ h=\frac{2.08\times {10}^{-19}}{0.3\times {10}^{15}}\\ =6.93\times {10}^{-34}\text{J}\cdot \text{s}\end{array}$

Thus, the value of the Planck’s constant is $6.93\times {10}^{-34}\text{J}\cdot \text{s}$.

Substitute $6.93\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (IV).

  $\begin{array}{c}\left(1.15\times {10}^{15}\right)\left(6.93\times {10}^{-34}\text{J}\cdot \text{s}\right)\text{J}=\varphi +\left(1.6\times {10}^{-19}\text{J}\right)\\ \varphi =\left(7.97\times {10}^{-19}\right)-\left(1.6\times {10}^{-19}\right)\\ =6.37\times {10}^{-19}\text{J}\end{array}$

Thus, the value of the Planck’s constant is $6.93\times {10}^{-34}\text{J}\cdot \text{s}$ and the value of the work function of the metal is $6.37\times {10}^{-19}\text{J}$.