Question

Photons with energy $5eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photoelectrons is $2eV$. When photons of energy $6eV$ are incident on $C$, no photoelectrons will reach the anode $A$, if the stopping potential of $A$ relative to $C$ is:

• A. $-3V$
• B. $+3V$
• C. $+4V$
• D. $-1V$

Answer 1

The correct option is A

$-3V$

Step 1: Given

First energy of photons emitted: $E_1=5eV$

Maximum energy of photoelectrons emitted: $KE=2eV$

Second energy of photons emitted: $E_2=6eV$

Step 2: Formula Used

1. The photoelectric equation is given as $E=\varphi +KE$, where $E$ is the frequency of photons incident, $\varphi$ is the work function of the metal and $KE$ is the maximum kinetic energy of emitted photoelectrons.
2. And, $KE=e{V}_0$, where $e$ is the charge of electrons and $V_0$ is the stopping potential.

Step 3: Calculate the work function of the metal.

Calculate the work function of the metal.

$$\begin{gathered} E=\varphi +KE \\ \Rightarrow 5eV=\varphi +2eV \\ \Rightarrow \varphi =5eV-2eV \\ \Rightarrow \varphi =3eV \end{gathered}$$

Step 4: Calculate the kinetic energy when the energy of photons emitted is $6eV$.

$$\begin{gathered} E=\varphi +KE \\ \Rightarrow 6eV=3eV+KE \\ \Rightarrow KE=6eV-3eV \\ \Rightarrow KE=3eV \end{gathered}$$

Step 5: Calculate the stopping potential.

$$\begin{gathered} KE=e{V}_0 \\ \Rightarrow 3eV=e{V}_0 \\ \Rightarrow V_0=3V \end{gathered}$$

Stopping potential is defined as the negative potential applied to the anode with respect to the cathode. Therefore, stopping potential of $A$ with respect to $C$ is $-3V$.

Hence, option A is correct.