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 ?

Options

A. `(7hf)/(12e)`
B. `(9hf)/(12e)`
C. `(29hf)/(12e)`
D. `(31hf)/(12e)`

Answer

A

Explanation

`phi =hf//2 " "phi_0 = hf//3`

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 conection conservind charge C is the capacitance of the capacitions and V is the common potential
`CV+CV=CV_(1) +CV_(2)`
`2V = (hf)/(2e) + (2hf)/(3e) rArr V = (7hf)/(12e)`