Question

What is the stopping potential when $\mathbf{250}\mathbf{}\mathit{n}\mathit{m}$ light strikes a zinc plate?

Short answer

The minimal negative voltage provided to the anode to halt the photocurrent is known as the stopping potential.The required stopping potential is $0.66V$ .

Step-by-step solution

A concept.

The energy of the photon is,

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$

Here, $\mathit{E}$is the energy of the photon, $\mathit{h}$is the plank’s constant, $\mathit{c}$ is the speed of light, and $\mathit{\lambda }$is the wavelength

Consider the given data as below.

Wavelength, $\lambda =250mm$

Plank’s constant, $h=4.135\times {10}^{-15}\frac{eV}{Hz}$

Speed of light, role="math" localid="1657549039174" $c=3\times {10}^8\frac{m}{s}=3\times {10}^{1}0\frac{cm}{s}$

Calculate the energy of the photon by substituting all known values into equation (1).

$$\begin{gathered} E=\frac{4.135\times {10}^{-15}\times 3\times {10}^{10}}{250} \\ =\frac{1240}{250} \\ =4.96eV \end{gathered}$$

Define the stopping potential.

The Kinetic energy of the photon from the zinc plate is given by:

${\mathbf{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{E}\mathbf{-}\mathit{\varphi }$

Here,$\mathit{\varphi }$ is the work function.

Consider the given data as below.

The work function of the material is $\varphi =4.6eV$.

Now the kinetic energy calculation,

$$\begin{gathered} {\mathrm{K}}_{\max }=\mathrm{E}-\varphi \\ =4.96-4.6 \\ =0.66eV \end{gathered}$$

 Define the stopping potential 

The stopping potential is given by,

$$\begin{gathered} eV=K{E}_{\max } \\ V=\frac{K{E}_{\max }}{e} \\ =\frac{0.66eV}{e} \\ =0.66V \end{gathered}$$

The stopping potential is given by,

Thus, the stopping potential is$=0.66V$.