Ultraviolet light by wavelength 200 nm is incident on the polished surface of Fe(Iron). The work function of the surface is 4.71 eV. What will be its stopping potential ?
`(h = 6.626xx10^(-34) J s, 1 eV = 1.6 xx10^(-19)J,)`

Options

A. 1.5 V
B. 2.5 V
C. 0.5 V
D. none of these

Answer

To solve the problem, we need to find the stopping potential (V₀) when ultraviolet light of wavelength 200 nm is incident on the polished surface of iron, which has a work function (φ) of 4.71 eV.

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
The wavelength (λ) is given as 200 nm. We need to convert this to meters:
\[ \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \]

**Hint**: Remember that 1 nm = \(10^{-9}\) m.

2. **Calculate Energy of Incident Photon**:
The energy (E) of the photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h = 6.626 \times 10^{-34} \, \text{J s}\) (Planck's constant)
- \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light)

Plugging in the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{200 \times 10^{-9} \, \text{m}} \]
\[ E = \frac{1.9878 \times 10^{-25}}{200 \times 10^{-9}} = 9.939 \times 10^{-19} \, \text{J} \]

**Hint**: Make sure to multiply and divide carefully, keeping track of units.

3. **Convert Energy from Joules to Electron Volts**:
To convert energy from Joules to electron volts (eV), use the conversion \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E = \frac{9.939 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 6.211 \, \text{eV} \]

**Hint**: Remember to divide the energy in Joules by the conversion factor to get the energy in eV.

4. **Calculate Kinetic Energy of Ejected Electrons**:
The kinetic energy (K.E.) of the ejected electrons can be calculated using the formula:
\[ K.E. = E - \phi \]
where φ is the work function of iron (4.71 eV):
\[ K.E. = 6.211 \, \text{eV} - 4.71 \, \text{eV} = 1.501 \, \text{eV} \]

**Hint**: The kinetic energy is the difference between the energy of the photon and the work function.

5. **Relate Kinetic Energy to Stopping Potential**:
The stopping potential (V₀) is related to the kinetic energy by:
\[ K.E. = eV_0 \]
where \(e\) is the charge of an electron (1 eV corresponds to 1 eV of energy per unit charge). Thus:
\[ V_0 = K.E. \]
Therefore:
\[ V_0 = 1.501 \, \text{V} \]

**Hint**: The stopping potential is numerically equal to the kinetic energy in eV.

### Final Answer:
The stopping potential \(V_0\) is approximately **1.50 V**.