Light of wavelength 200 nm incident on a metal surface of threshold wavelength 400 nm kinetic energy of fastest photoelectron will be

Options

A. `3.1 eV`
B. `4.5 eV`
C. `6.2 eV`
D. `7.2 eV`

Answer

To find the kinetic energy of the fastest photoelectron when light of wavelength 200 nm is incident on a metal surface with a threshold wavelength of 400 nm, we can follow these steps:

### Step 1: Understand the photoelectric effect
The photoelectric effect states that when light of sufficient energy (or short enough wavelength) strikes a metal surface, it can eject electrons from that surface. The energy of the incident photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\(6.63 \times 10^{-34} \, \text{J s}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength of the light.

### Step 2: Calculate the energy of the incident photon
Given that the wavelength of the incident light \( \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \):
\[ E_{\text{incident}} = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{200 \times 10^{-9} \, \text{m}} \]

### Step 3: Calculate the threshold energy
The threshold wavelength \( \lambda_0 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \):
\[ E_{\text{threshold}} = \frac{hc}{\lambda_0} = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \]

### Step 4: Calculate the kinetic energy of the photoelectron
The kinetic energy \( KE \) of the photoelectron is given by the difference between the energy of the incident photon and the threshold energy:
\[ KE = E_{\text{incident}} - E_{\text{threshold}} \]

### Step 5: Substitute the values and calculate
1. Calculate \( E_{\text{incident}} \):
\[ E_{\text{incident}} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{200 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{200 \times 10^{-9}} = 9.945 \times 10^{-19} \, \text{J} \]

2. Calculate \( E_{\text{threshold}} \):
\[ E_{\text{threshold}} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{400 \times 10^{-9}} = 4.9725 \times 10^{-19} \, \text{J} \]

3. Calculate \( KE \):
\[ KE = 9.945 \times 10^{-19} - 4.9725 \times 10^{-19} = 4.9725 \times 10^{-19} \, \text{J} \]

### Step 6: Convert kinetic energy to electron volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE = \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV} \]

### Final Answer
The kinetic energy of the fastest photoelectron is approximately **3.1 eV**.

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