Question

A photon having energy \(5.5 \mathrm{eV}\) is incident on metal suface and emits photo electrons having maximum kinetic energy \(0.4 \mathrm{eV}\). Then stopping potential of this electron is..... (A) \(5.5 \mathrm{eV}\) (B) \(5.1 \mathrm{eV}\) (C) \(5.9 \mathrm{eV}\) (D) \(0.4 \mathrm{eV}\)

Short answer

The stopping potential of the emitted electrons is \(0.4 \mathrm{eV}\). The correct option is (D).

Step-by-step solution

Understand the photoelectric effect and given quantities

The photoelectric effect is observed when photons of sufficient energy are incident upon a metal surface, resulting in the emission of electrons. These electrons are called 'photoelectrons'. In this exercise, we are given the energy of the incident photon (5.5 eV) and the maximum kinetic energy (0.4 eV) of the emitted electrons.

Find the work function of the metal

The work function denoted by \(\phi\) is the minimum energy required to eject an electron from the metal surface. According to Einstein's photoelectric equation, the energy of the incident photon equals the sum of the maximum kinetic energy of the electron and the work function of the metal: \[E_{photon} = K_{max} + \phi\] We can rearrange this equation to find the work function: \[\phi = E_{photon} - K_{max}\] Substitute the given energy values into the equation: \[\phi = 5.5 \mathrm{eV} - 0.4 \mathrm{eV}\] \[\phi = 5.1 \mathrm{eV}\]

Calculate the stopping potential

The stopping potential (V) is the potential required to stop the emitted electrons from reaching the anode in a photoelectric cell. It is related to the work function and the energy of the incident photons by the following equation: \[E_{photon} = eV + \phi\] Where e is the charge of an electron (\(1.6 \times 10^{-19} C\)). Rearrange the equation to find the stopping potential: \[V = \frac{E_{photon} - \phi}{e}\] We have already calculated the work function in Step 2. So, substitute the work function and incident photon energy values into the equation: \[V = \frac{5.5 \mathrm{eV} - 5.1 \mathrm{eV}}{1.6 \times 10^{-19} C}\] \[V = \frac{0.4 \mathrm{eV}}{1.6 \times 10^{-19} C}\] Since 1 eV = \(1.6 \times 10^{-19} C\), we can calculate the stopping potential in volts: \[V = 0.4 \mathrm{V}\] This is the same as 0.4 eV, as 1 eV = 1 V for a one-electron charge. So the stopping potential of the emitted electrons is 0.4 eV. The correct option is (D).