Work function of a cathode is 4.2 eV. Energy of incident photons is `6.2 eV `. Anode is kept at ` + 2 ` volt w.r.t cathode. Calculate minimum and maximum K.E. of electrons reaching at the anode.

Answer

To solve the problem, we need to calculate the minimum and maximum kinetic energy (K.E.) of electrons reaching the anode after being emitted from the cathode due to the photoelectric effect.

### Given Data:
- Work function of the cathode (Φ) = 4.2 eV
- Energy of incident photons (E_photon) = 6.2 eV
- Anode potential (V_anode) = +2 V with respect to cathode

### Step 1: Calculate the Kinetic Energy of Electrons Emitted from the Cathode
Using the formula for the kinetic energy of emitted electrons:
\[ K.E. = E_{photon} - \Phi \]
Substituting the values:
\[ K.E. = 6.2 \, \text{eV} - 4.2 \, \text{eV} = 2.0 \, \text{eV} \]

### Step 2: Determine the Minimum Kinetic Energy at the Anode
The minimum kinetic energy of the electrons reaching the anode occurs when the electrons have the least initial kinetic energy (which is 0 eV). When these electrons reach the anode, they gain additional energy due to the potential difference between the cathode and the anode.

The energy gained due to the potential difference is given by:
\[ \text{Energy gained} = e \cdot V_{anode} = 1 \cdot 2 \, \text{eV} = 2 \, \text{eV} \]
Thus, the minimum kinetic energy at the anode is:
\[ K.E._{min} = 0 \, \text{eV} + 2 \, \text{eV} = 2 \, \text{eV} \]

### Step 3: Determine the Maximum Kinetic Energy at the Anode
The maximum kinetic energy of the electrons reaching the anode occurs when the electrons have the maximum kinetic energy emitted from the cathode (which we calculated as 2 eV). These electrons also gain energy from the potential difference.

Thus, the maximum kinetic energy at the anode is:
\[ K.E._{max} = 2 \, \text{eV} + 2 \, \text{eV} = 4 \, \text{eV} \]

### Final Results:
- Minimum Kinetic Energy (K.E._min) = 2 eV
- Maximum Kinetic Energy (K.E._max) = 4 eV

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