Question

When a metal was exposed to light at a frequency of 3.64× 1015 s–1, electrons were emitted with a kinetic energy of 5.80× 10–19 J.What is the maximum number of electrons that could be ejected from this metal by a burst of light (at some other frequency) with a total energy of 8.66× 10–7 J?

Answer 1

Answer :  The maximum number of electrons released $4.73\times 10^{12}electrons$

Explanation : Given,

Frequency = $3.64\times 10^{15}s^{-1}$

Kinetic energy = $5.80\times 10^{-19}J$

Total energy = $8.66\times 10^{-7}J$

First we have to calculate the work function of the metal.

Formula used :

$K.E=h\nu -w$

where,

K.E = kinetic energy

h = Planck's constant = $6.626\times 10^{-34}J/s$

$\nu$ = frequency  

w = work function

Now put all the given values in this formula, we get the work function of the metal.

$5.80\times 10^{-19}J=(6.626\times 10^{-34}J/s\times 3.64\times 10^{15}s^{-1})-w$

By rearranging the terms, we get

$w=1.83\times 10^{-18}J$

Therefore, the works function of the metal is, $1.83\times 10^{-18}J$

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released = $\frac{\text{ Total energy}}{\text{ work function}}$

$\frac{8.66\times 10^{-7}J}{1.83\times 10^{-19}J}=4.73\times 10^{12}electrons$

Therefore, the maximum number of electrons released is $4.73\times 10^{12}electrons$