Question

In photoelectric experiment energy of 2.48 eV irradiates a photo sensitive material. The stopping potential was measured to be 0.5 V. Work function of the photo sensitive material is :

• A. 0.5 eV
• B. 1.68 eV
• C. 2.48 eV
• D. 1.98 eV

Answer 1

The Correct Option is D

The energy relation for the photoelectric effect is:

\[ eV_s = h\nu - \phi \]

where $eV_s$ is the stopping potential energy, $h\nu$ is the energy of incident photons, and $\phi$ is the work function.

Given: $h\nu = 2.48 \, \text{eV}, \, V_s = 0.5 \, \text{V}.$

\[ 0.5 = 2.48 - \phi \]

\[ \phi = 2.48 - 0.5 = 1.98 \, \text{eV}. \]

The work function of the material is $\phi = 1.98 \, \text{eV}.$

Answer 2

This problem is based on the photoelectric effect. We are given the energy of the incident photons and the stopping potential required to stop the photoelectrons. We need to calculate the work function of the photosensitive material.

Concept Used:

The solution uses Einstein's photoelectric equation, which describes the conservation of energy in the photoelectric effect. The energy of an incident photon (\(E\)) is used in two ways: a part of it is used to overcome the binding energy of the electron in the metal (the work function, \(\phi\)), and the remaining energy is converted into the kinetic energy of the emitted photoelectron (\(K_{max}\)).

The equation is:

\[ E = \phi + K_{max} \]

The maximum kinetic energy (\(K_{max}\)) of the photoelectrons is related to the stopping potential (\(V_s\)) by the equation:

\[ K_{max} = e V_s \]

where \(e\) is the elementary charge. When \(V_s\) is in Volts (V), the kinetic energy \(K_{max}\) can be conveniently expressed in electron-volts (eV).

Step-by-Step Solution:

Step 1: Identify the given values from the problem statement.

Energy of the incident photons, \(E = 2.48 \text{ eV}\).

Stopping potential, \(V_s = 0.5 \text{ V}\).

Step 2: Calculate the maximum kinetic energy (\(K_{max}\)) of the photoelectrons using the stopping potential. The relationship is \(K_{max} = eV_s\).

\[ K_{max} = e \times (0.5 \text{ V}) \]

This gives the kinetic energy directly in electron-volts:

\[ K_{max} = 0.5 \text{ eV} \]

Step 3: Use Einstein's photoelectric equation to find the work function (\(\phi\)). The equation is \(E = \phi + K_{max}\).

Rearranging the formula to solve for the work function:

\[ \phi = E - K_{max} \]

Step 4: Substitute the known values of \(E\) and \(K_{max}\) into the rearranged equation.

\[ \phi = 2.48 \text{ eV} - 0.5 \text{ eV} \]

Final Computation & Result:

Performing the subtraction gives the value of the work function.

\[ \phi = 1.98 \text{ eV} \]

Therefore, the work function of the photo sensitive material is 1.98 eV.