Light from a hydrogen tube is incident on the cathode of a photoelectric cell the work function of the cathode surface is `4.2 eV`. In order to reduce the photo - current to zero the voltage of the anode relative to the cathode must be made

Options

A. `-4.2 V`
B. `-9.4 V`
C. `-17.8 V`
D. `+9.4 V`

Explanation

To solve the problem, we need to determine the stopping potential required to reduce the photocurrent to zero when light from a hydrogen tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is given as 4.2 eV.

### Step-by-Step Solution:

1. **Identify the Energy of the Incident Light:**
The energy of the incident photon from the hydrogen tube can be calculated using the formula for the energy levels of hydrogen:
\[ E = -\frac{13.6 \, \text{eV}}{n^2} \]
For the ground state (n = 1), the energy is:
\[ E = -13.6 \, \text{eV} \quad \text{(this is the energy required to remove the electron)} \]
However, when considering the energy of the photon that can eject an electron, we take the positive value:
\[ E_{\text{photon}} = 13.6 \, \text{eV} \]

2. **Calculate the Maximum Kinetic Energy of Ejected Electrons:**
The maximum kinetic energy (K.E.) of the photoelectrons can be calculated using the equation:
\[ K.E. = E_{\text{photon}} - \phi \]
where \(\phi\) is the work function of the cathode. Given \(\phi = 4.2 \, \text{eV}\):
\[ K.E. = 13.6 \, \text{eV} - 4.2 \, \text{eV} = 9.4 \, \text{eV} \]

3. **Determine the Stopping Potential:**
To reduce the photocurrent to zero, we need to apply a stopping potential (\(V_0\)) that will counteract the kinetic energy of the ejected electrons. The relationship between the loss in kinetic energy and the gain in potential energy is given by:
\[ K.E. = e \cdot V_0 \]
where \(e\) is the charge of the electron (which cancels out in the equation). Thus, we can set:
\[ 9.4 \, \text{eV} = e \cdot V_0 \]
Therefore, the stopping potential \(V_0\) required is:
\[ V_0 = -9.4 \, \text{V} \]
The negative sign indicates that the potential is applied in the opposite direction to the motion of the electrons.

4. **Conclusion:**
The voltage of the anode relative to the cathode must be made \(-9.4 \, \text{V}\) to reduce the photocurrent to zero.

### Final Answer:
The voltage of the anode relative to the cathode must be made **-9.4 V**.

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