Question

A metal of work function 2.3 eV is irradiated with radiation of wavelength \( y \times 10^2 \) nm. The maximum kinetic energy of ejected electron is \( 2.8 \times 10^{-20} \) J. Then calculate \( y \) (in nearest integer).

Hint:

In the photoelectric effect, the energy of the ejected electron is the difference between the energy of the incident photon and the work function of the material.

Answer 1

Correct Answer: 5

Step 1: Use the photoelectric equation.
The photoelectric equation is given by: \[ E_{\text{photon}} = \text{Work Function} + \text{Kinetic Energy of ejected electron} \] where the energy of the photon is given by: \[ E_{\text{photon}} = \frac{h c}{\lambda} \] Here, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)), \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength of the radiation.
Step 2: Convert work function to joules.
Given: \[ \text{Work Function} = 2.3 \, \text{eV} = 2.3 \times 1.602 \times 10^{-19} \, \text{J} \]
Step 3: Apply the given values.
The maximum kinetic energy of the electron is given as: \[ KE = 2.8 \times 10^{-20} \, \text{J} \] Thus, the energy of the photon is: \[ E_{\text{photon}} = 2.3 \times 1.602 \times 10^{-19} + 2.8 \times 10^{-20} \] \[ E_{\text{photon}} = 3.68 \times 10^{-19} \, \text{J} \]
Step 4: Solve for wavelength \( \lambda \).
Using the formula \( E_{\text{photon}} = \frac{h c}{\lambda} \), we get: \[ \lambda = \frac{h c}{E_{\text{photon}}} \] Substituting the known values: \[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) (3 \times 10^8 \, \text{m/s})}{3.68 \times 10^{-19} \, \text{J}} \] \[ \lambda = 5.4 \times 10^{-7} \, \text{m} = 540 \, \text{nm} \]
Step 5: Final Answer.
Thus, \( y = 5 \).