Question

In photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelength of incident light is 260nm, the maximum kinetic energy electrons will be:
Given $E\left( {{\text{in eV}}} \right) = \dfrac{{1237}}{{\lambda ({\text{in nm)}}}}$
(A) $1.5eV$
(B) $4.5eV$
(C) $15.1eV$
(D) $3.0eV$

Answer 1

The maximum kinetic energy of ejected electrons in a photoelectric experiment can be given by the energy of the radiation minus the work function of the metal. The work function is an energy threshold which is the minimum energy required to eject the electron without giving it a kinetic energy.
Formula used: In this solution we will be using the following formulae;
$K{E_{\max }} = E - {E_0}$ where $K{E_{\max }}$ is the maximum kinetic energy of an electron, $E$ is the energy of the incident radiation, and ${E_0}$ is the work function (threshold energy) of the metal.

Complete Step-by-Step Solution:
We are told that a light of wavelength of 260 nm used as an incident light in a photoelectric experiment. Whatever the metal that was used, we are informed that the light must have a threshold wavelength of 380 nm (maximum wavelength which will eject an electron). We are to determine the maximum kinetic energy of the electrons.
We note that the kinetic energy can be given as
$K{E_{\max }} = E - {E_0}$ where $K{E_{\max }}$ is the maximum kinetic energy of an electron, $E$ is the energy of the incident radiation, and ${E_0}$ is the work function (threshold energy) of the metal. This can be written as
$K{E_{\max }} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$ since $E = \dfrac{{hc}}{\lambda }$ where $h$ is Planck’s constant and $c$ is speed of light, $\lambda$ is the wavelength.
As given, $E\left( {{\text{in eV}}} \right) = \dfrac{{1237}}{{\lambda ({\text{in nm)}}}}$
Then,
$K{E_{\max }} = \dfrac{{1237}}{{260}} - \dfrac{{1237}}{{380}}$
By computation, we have
$K{E_{\max }} = 1.5eV$

Hence, the correct option is A

Note: Although it may be odd or confusing, observe that the minimum energy to eject a photon corresponds to the maximum wavelength required. This is because the wavelength and energy are inversely related.