A monochromatic light of frequency f is incident on two identical metal spheres of threshold frequency `f/2` and `f/3` respectively. After some time, emission of photoelectron will stop on both spheres . Now both metal spheres are connected through wire. (radius of spherhes is R )
(a)What will be potential of spheres now ?
(b)how many electron will flow through wire ?

Answer

(i) `V' = (7hf)/(12)` (ii) `Deltan = (hfR)/(12eK)`

Explanation

Potential at which electron stop coming out
From sphere-1, `V_(1) = ((hf-(hf)/(2)))/(e) = (hf)/(2e)`
from sphere-2, `V_(2) = (hf-(hf)/(3))/(e) = (2hf)/(3e)`
After connection
(i) `V + V = V_(1) + V_(2)` (V = final common potential)
`rArr 2V = hf/2e + (2hf)/(3e) rArr V = (7hf)/(12e)`
(ii) For sphere-2 : `(kDeltaR)/(R) = ((2/3(hf)/(e)) - (7hf)/(12e)) = (hf)/(12e)`
No. of electrons flows `Deltan = (DeltaQ)/(e) = (hfR)/(12ke^(12))`