The photo electric threshold wavelength of Tungsten is 2300 Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelenghth 1800 Å is

Options

A. 0.15 eV
B. 1.5 eV
C. 15 eV
D. 150 eV

Explanation

To solve the problem, we need to calculate the energy of the electrons ejected from the surface of tungsten when exposed to ultraviolet light of wavelength 1800 Å. We will follow these steps:

### Step 1: Calculate the Work Function (Φ) of Tungsten
The work function (Φ) can be calculated using the formula:
\[ \Phi = \frac{hc}{\lambda_0} \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda_0\) is the threshold wavelength (2300 Å = \(2300 \times 10^{-10} \, \text{m}\)).

Substituting the values:
\[ \Phi = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{2300 \times 10^{-10} \, \text{m}} \]
Calculating this gives:
\[ \Phi \approx 5.39 \, \text{eV} \]

### Step 2: Calculate the Energy of the Incident Photon
The energy of the incident photon can be calculated using the same formula:
\[ E = \frac{hc}{\lambda} \]
where \(\lambda\) is the wavelength of the incident light (1800 Å = \(1800 \times 10^{-10} \, \text{m}\)).

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{1800 \times 10^{-10} \, \text{m}} \]
Calculating this gives:
\[ E \approx 6.88 \, \text{eV} \]

### Step 3: Calculate the Kinetic Energy of the Ejected Electrons
The kinetic energy (KE) of the ejected electrons can be calculated using the equation:
\[ KE = E - \Phi \]
Substituting the values we found:
\[ KE = 6.88 \, \text{eV} - 5.39 \, \text{eV} \approx 1.49 \, \text{eV} \]
This can be approximated to:
\[ KE \approx 1.5 \, \text{eV} \]

### Final Answer
The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800 Å is approximately **1.5 eV**.

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