Question

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5\, V$ when the radiation of $250\, nm$ is used. The work function of the metal is :

• A. 4 eV
• B. 4.5 eV
• C. 5 eV
• D. 5.5 eV

Answer 1

The Correct Option is B

To solve this problem, we start by understanding the photoelectric effect, where electrons are ejected from a metal surface when it is exposed to light of sufficient frequency.

The key formula for the photoelectric effect is:

E_k = h\nu - \phi

Where:

• E_k is the kinetic energy of the ejected photoelectron
• h is Planck’s constant: 6.626 \times 10^{-34} \, \text{J s}
• \nu is the frequency of the incident radiation
• \phi is the work function of the metal

Given:

• Stopping potential V_s = 0.5 \, \text{V}
• Wavelength \lambda = 250 \, \text{nm} = 250 \times 10^{-9} \, \text{m}

The relationship between stopping potential and kinetic energy is:

E_k = eV_s

where e is the elementary charge 1.602 \times 10^{-19} \, \text{C}.

Now we calculate the energy of the incident photon using:

E_{\text{photon}} = \frac{hc}{\lambda}

Where:

• c is the speed of light: 3 \times 10^8 \, \text{m/s}

Substituting the values:

E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{250 \times 10^{-9}}

Calculating gives:

E_{\text{photon}} \approx 7.95 \times 10^{-19} \, \text{J}

Convert this energy from joules to electronvolts (1 eV = 1.602 \times 10^{-19} \, \text{J}):

E_{\text{photon}} \approx \frac{7.95 \times 10^{-19}}{1.602 \times 10^{-19}} \, \text{eV} \approx 4.96 \, \text{eV}

Now calculate the work function:

\phi = E_{\text{photon}} - eV_s

Where eV_s = 0.5 \, \text{eV}. Therefore:

\phi = 4.96 \, \text{eV} - 0.5 \, \text{eV} = 4.46 \, \text{eV} \approx 4.5 \, \text{eV}

Thus, the work function of the metal is approximately 4.5 eV.