Question

Photoelectric effect takes place in element A. Its work function is $2.5eV$ and threshold wavelength is $\lambda$. Another element B is having a work function of $5eV$. Then find out the maximum wavelength that can produce photoelectric effect in B.
(A) $\dfrac{\lambda }{2}$
(B) $2\lambda$
(C) $\lambda$
(D) $3\lambda$

Answer 1

The work function of two elements are given. The threshold wavelength of element a is given we have to find the maximum wavelength that can produce photoelectric effect in B. Find the work function of elements A and B using Einstein's photoelectric equation.

Complete step by step solution:
Einstein conducted experiments on the photons and formed the photoelectric equation that gives the energy of the photon. Photons are allowed to hit on metal surfaces, when the photon hits the surface of the metal electron from the metal surface it gets ejected, it is known as the photoelectron using this mechanism photoelectric current is produced . Based on this experiment, the energy of the incident photon is calculated.
The incident energy of the photon is equal to the sum of work function and the kinetic energy of the photon.
$\Rightarrow {\text{I}}{\text{.E = }}\Phi {\text{ + K}}{\text{.E}}$
$\Rightarrow \Phi = \dfrac{{I.E}}{{K.E}}{\text{ }} \to {\text{1}}$
I.E is the energy of incident electron
$\Phi$ is the work function of the metal surface
K.E is the kinetic energy of the incident photon
$\Rightarrow I.E = hf$
$\Rightarrow K.E = \dfrac{1}{2}m{v^2}$
h is the Planck’s constant
V is the velocity
m is the mass
f is the frequency
The equation 1 becomes
$\Rightarrow \Phi = \dfrac{{hf{\text{ }}}}{{\dfrac{1}{2}m{v^2}}}{\text{ }} \to {\text{2}}$
We know that frequency is inversely proportional to wavelength
$\Rightarrow \nu = \dfrac{1}{\lambda }$
Then equation 2 becomes
$\Rightarrow \Phi = \dfrac{{2{\text{ }}}}{{\lambda m{v^2}}}$
Given,
Work function of the plate A is ${\Phi _1} = 2.5eV$
$\Rightarrow {\Phi _1} = \dfrac{{2{\text{ }}}}{{{\lambda _1}m{v^2}}}$
$\Rightarrow {\lambda _1} = \dfrac{{2{\text{ }}}}{{2.5m{v^2}}}$
Work function of the plate B is ${\Phi _1} = 5eV$
$\Rightarrow {\Phi _2} = \dfrac{{2{\text{ }}}}{{{\lambda _2}m{v^2}}}$
$\Rightarrow {\lambda _2} = \dfrac{{2{\text{ }}}}{{5m{v^2}}}$
If we divide the work function of plate A by plate B
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\dfrac{{2{\text{ }}}}{{2.5m{v^2}}}}}{{\dfrac{{2{\text{ }}}}{{5m{v^2}}}}}$
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{\text{5 }}}}{{2.5}}$
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = 2$
$\Rightarrow {\lambda _2} = \dfrac{{{\lambda _1}}}{2}$
The threshold wavelength of element 1 is $\lambda$. So,
$\Rightarrow {\lambda _2} = \dfrac{\lambda }{2}$
The maximum wavelength that can produce photoelectric effect in B, ${\lambda _{\max }} = \dfrac{\lambda }{2}$

Hence the correct answer is option A) $\dfrac{\lambda }{2}$

Note: Work function is defined as the amount of energy required to eject an electron from the metal surface. If the energy of the incident electron is less than the work function then the photoelectron will not be ejected from the metal surface.