Question

The electric field of a light wave is given as, $\gamma \dfrac{N}{C}$. This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is:
Given E (in eV) =12375/ λ (in A˚ )
A. 0.48V
B. 2.0V
C. 2.48V
D. 0.72V

Answer 1

Hint: Einstein's equation can be used to solve this question. The kinetic energy, wavelength should be found out meanwhile. The unit of kinetic energy is joule and that of wavelength is meter.

Complete step by step answer:
First of all let us look at the definition of stopping potential. Stopping potential is the voltage difference required to stop the flow of electrons in a material.
That is, the minimum potential required to stop the flow of electrons.
We all know that Einstein’s equation,
$K{{E}_{\max }}=E-\phi $
Angular frequency is given by,
Since we know that frequency,
 And the wavelength can be found out using the formula,
$\lambda =\dfrac{c}{f}$
Applying the values of velocity of light and the frequency of the wave gives the wavelength of light
That is,
We also knows that
 $\lambda =\dfrac{3\times {{10}^{8}}}{6\times {{10}^{^{14}}}}$
Calculating the value of wavelength
$\lambda =5000$Angstrom
Therefore after further solving,
Then,
$e{{V}_{s}}=0.475eV$
Both the sides are having electronic charges so that we can cut or remove it from both the sides.
Therefore now we get,
${{V}_{s}}=0.475V$
In short, by approximation we get,
V=0.48V

Therefore it is found that option A is the correct answer.

Note:
The usage of exponential may lead to faulty answers taking care of that. And also take care of the units and dimensions of the quantities like wavelength mentioned in the question. Einstein’s equation can be used to solve this question. The kinetic energy, wavelength should be found out meanwhile.