Question

Light of wavelength 5000 $A^\circ $ falls on a sensitive plate with photoelectric work function 1.9eV. The maximum kinetic energy of the photoelectrons emitted will be:
1) 0.58eV
2) 2.48eV
3) 1.24eV
4) 1.16eV

Answer 1

Hint: Work function or binding energy is defined as the minimum amount of energy which is required to remove an electron to infinity from its source. It was defined in the concept of photoelectric effect in which a light is incident upon the metal and the electrons are ejected from the metal.

Complete step by step solution:
Here the formula for maximum kinetic energy is:

$K{E_{\max }} = hf - BE$;
Here:

\[K{E_{\max }}\]= Maximum kinetic energy.

h = Planck’s Constant;
f = frequency;
BE = Binding energy or work function;
Find out the frequency of the light:
We know the speed of light is $3 \times {10^8}m/s$.
 Formula for frequency is given by:

$f = \dfrac{c}{\lambda }$ ;

Here:
f = frequency;
c = Speed of light;

$\lambda $= Wavelength;

Put the above relation of frequency in the formula for K.E:

$K{E_{\max }} = \dfrac{{hc}}{\lambda } - BE$;

Put the given value in the above equation and solve:

$K{E_{\max }} = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5000 \times {{10}^{ - 10}}}} - 1.9$;

Convert the numerator which is in joules to electron volt:

$ \Rightarrow K{E_{\max }} = \dfrac{{12375}}{{5000}} - 1.9$;
$ \Rightarrow K{E_{\max }} = 2.475 - 1.9$;

The maximum K.E. will be:

$ \Rightarrow K{E_{\max }} = 0.575eV$;
$ \Rightarrow K{E_{\max }} \simeq 0.58eV$;

Final Answer: Option “1” is correct. Light of wavelength 5000 $A^\circ $falls on a sensitive plate with photoelectric work function 1.9eV. The maximum kinetic energy of the photoelectrons emitted will be$0.58eV$.

Note: Here, we have to first write the formula for the maximum kinetic energy in terms of work function or binding energy and we have to write the frequency in terms of velocity of light and wavelength and put in the formula for K.E. put the given value in the formula and find the unknown.