The photoelectric threshold for some material is 200 nm. The material is irradiated with radiations of wavelength 40 nm. The maximum kinetic energy of the emitted photoelectrons is

Options

A. 2 eV
B. 1 eV
C. 0.5 eV
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy of the incident photons
We can use the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\) (Planck's constant)
- \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light)
- \(\lambda = 40 \, \text{nm} = 40 \times 10^{-9} \, \text{m}\) (wavelength of the incident light)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{40 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E = \frac{1.9878 \times 10^{-25}}{40 \times 10^{-9}} = 4.9695 \times 10^{-18} \, \text{J} \]

### Step 2: Convert the energy from Joules to electron volts
To convert energy from Joules to electron volts, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ E = \frac{4.9695 \times 10^{-18} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 31.18 \, \text{eV} \]

### Step 3: Calculate the work function (\(W\)) of the material
The work function is given by:
\[ W = \frac{hc}{\lambda_0} \]
where \(\lambda_0 = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m}\).

Substituting the values:
\[ W = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{200 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ W = \frac{1.9878 \times 10^{-25}}{200 \times 10^{-9}} = 9.939 \times 10^{-19} \, \text{J} \]

### Step 4: Convert the work function from Joules to electron volts
\[ W = \frac{9.939 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 6.21 \, \text{eV} \]

### Step 5: Calculate the maximum kinetic energy of the emitted photoelectrons
Using Einstein's photoelectric equation:
\[ KE_{\text{max}} = E - W \]
Substituting the values:
\[ KE_{\text{max}} = 31.18 \, \text{eV} - 6.21 \, \text{eV} = 24.97 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the emitted photoelectrons is approximately:
\[ KE_{\text{max}} \approx 25 \, \text{eV} \]