A student studying the photoelectric effect from two different metals records the following information:

(i) the stopping potential for photoelectrons released from metal 1 is 1.48V larger than that for metal 2, and

(ii) the threshold frequency for metal 1 is 40.0% smaller than that for metal

2. Determine the work function for each metal.

What would be the final answer?

Details:

Given

stopping potential for metal-1 = 1.48 V

stopping potential for metal - 2 = V

Threshold frequency for metal -1 = λ- 0.4 λ = 0.6λ

Threshold frequency of metal -2 = λ

There fore the relation between stopping potential , threshold frequency and work function

eV = h f - φ

substitute the values in the above equation

so ,

For metal - 1

e (V + 1.48) = h c / 0.6 λ -φ₁

The work function of metal -1

φ₁ = hc / 0.6 λ - e (V+1.48 )

For metal - 2

e V = h c / λ - φ₂

The work function of metal -2

φ₂ = h c / λ - eV

where h = plancksconstant

= 6.63 x 10^-34 Js

c = speed of light

= 3 x 10⁸ m/s

e = charge of electron

= 1.6 x 10^-19 C