Question

Radiation containing frequencies \(6 \times 10^{14}\,\text{Hz}\) and \(9 \times 10^{14}\,\text{Hz}\) falls on a metal surface with work function \(2.5\,\text{eV}\). Calculate the maximum energy of emitted electrons (in eV).

• A. \(0.8\)
• B. \(1.0\)
• C. \(1.2\)
• D. \(1.5\)

Hint:

In photoelectric effect problems, always use the highest incident frequency to calculate maximum kinetic energy.

Answer 1

The Correct Option is C

To find the maximum energy of the emitted electrons when radiation containing frequencies \(6 \times 10^{14}\,\text{Hz}\) and \(9 \times 10^{14}\,\text{Hz}\) falls on a metal surface with a work function of \(2.5\,\text{eV}\), we will use the photoelectric effect equation:

\(E_{k} = h \nu - \phi\) 

where:

• \(E_{k}\) is the maximum kinetic energy of the emitted electrons (in eV).
• \(h\) is Planck's constant, \(4.1357 \times 10^{-15}\,\text{eV}\cdot\text{s}\).
• \(\nu\) is the frequency of the incident radiation.
• \(\phi\) is the work function of the metal (in eV).

We will calculate the kinetic energy for both frequencies and choose the maximum value:

1. For frequency \(6 \times 10^{14}\,\text{Hz}\): \(E_{k1} = h \nu_{1} - \phi = (4.1357 \times 10^{-15}\,\text{eV}\cdot\text{s}) \times (6 \times 10^{14}\,\text{Hz}) - 2.5\,\text{eV}\) \(= 2.48142\,\text{eV} - 2.5\,\text{eV} = -0.01858\,\text{eV}\)

The negative result implies that no electrons are emitted for this frequency, as the energy is less than the work function.

1. For frequency \(9 \times 10^{14}\,\text{Hz}\): \(E_{k2} = h \nu_{2} - \phi = (4.1357 \times 10^{-15}\,\text{eV}\cdot\text{s}) \times (9 \times 10^{14}\,\text{Hz}) - 2.5\,\text{eV}\) \(= 3.72213\,\text{eV} - 2.5\,\text{eV} = 1.22213\,\text{eV}\)

This indicates that the emitted electrons have a maximum kinetic energy of approximately \(1.2\,\text{eV}\).

Therefore, the maximum energy of the emitted electrons is \(1.2\,\text{eV}\), which matches the given correct answer.