Radiation from a hydrogen discharge discharge tube is incident on the cathode ofa photocell having energy 3.4 eV .The work function of the cathode suface is 3.2 eV. To reduce the photocurrent to zero, the voltage (in volts) of th anode realtive to the cathode must be made

Options

A. `+ 0.2`
B. `+3.2`
C. `+10.4`
D. `+13.6`

Explanation

To solve the problem step by step, we need to analyze the given information and apply the photoelectric effect principles.

### Step 1: Identify the given values
- Energy of the incident radiation (E) = 3.4 eV
- Work function of the cathode (φ) = 3.2 eV

### Step 2: Calculate the maximum kinetic energy (Kmax) of the emitted electrons
Using the photoelectric equation:
\[ K_{max} = E - φ \]
Substituting the values:
\[ K_{max} = 3.4 \, \text{eV} - 3.2 \, \text{eV} \]
\[ K_{max} = 0.2 \, \text{eV} \]

### Step 3: Relate the maximum kinetic energy to the stopping potential (V0)
The stopping potential (V0) is the potential required to stop the emitted electrons, which is equal to the maximum kinetic energy:
\[ K_{max} = e \cdot V_0 \]
Where e is the charge of an electron (1 eV corresponds to 1 volt when considering energy in eV). Thus:
\[ V_0 = K_{max} \]
\[ V_0 = 0.2 \, \text{V} \]

### Step 4: Conclusion
To reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:
\[ V_0 = 0.2 \, \text{V} \]

### Final Answer:
The voltage of the anode relative to the cathode must be made +0.2 V.

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