Question

A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stop vs ν is given in Figure.

1. Which material A or B has a higher work function?
2. Given the electric charge of an electron = 1.6 × 10^–19 C, find the value of h obtained from the experiment for both A and B.

Comment on whether it is consistent with Einstein’s theory:

Answer 1

Threshold frequency (no): The minimum frequency of incident radiations, required to eject the electron from the metal surface is defined as the threshold frequency.

If incident frequency n < n₀ ⇒ No photoelectron emission.

For most metals, the threshold frequency is in the ultraviolet (corresponding to wavelengths between 200 and 300 nm), but for potassium and caesium oxides it is in the visible spectrum (A between 400 and 700 nm).

Here we are given the threshold frequency of A

i. v_OA = 5 × 10¹⁴ Hz and 

For B, v_OB = 10 × 0¹⁴ Hz

We know that

Work function, `phi = hv_0` or `phi_0 ∝ v_0`

⇒ `phi_0 ∝ v_0`

So, `phi_(OA)/phi_(OB) = (5 xx 10^14)/(10 xx 10^14) < 1`

⇒ `phi_(OA) < phi_(OB)`

Thus, work function of B is higher than A.

ii. For metal A, slope = `h/e = 2/(10 - 15)10^14`

or `h = (2e)/(5 xx 10^14) = (2 xx 1.6 xx 10^-19)/(5 xx 10^14)`

= `6.4 xx 10^-34` Js

For metal B, sloe = `h/e = 2.5/((15 - 10)10^14`

or `h = (2.5 xx e)/(5 xx 10^14) = (2.5 xx 1.6 xx 10^-19)/(5 xx 10^14)`

= 8 × 10^–34 Js

Since the value of h from the experiment for metals A and B is different. Hence, the experiment is not consistent with the theory.