Question

In a photocell the plates P and Q have a separation of 5 cm, which are connected through a galvanometer without any cell. Bichromatic light of wavelengths `4000 Å` and `6000 Å` are incidenton plate Q whose work- functionis 2.39 eV. If a uniform magnetic field B exists parallel to the plates, find the minimum value of B for which the galvanometer shows zero deflection.

Answer 1

Correct Answer - A
Energy of photons corresponding to light of wavelength `lambda_1 = 4000 Å` is
`E_1 = (12375)/(4000) = 3.1 eV`
and that corresponding to `lambda_2 = 6000 Å is `
`E_2 = (12375)/(6000) = 2.06 eV`
As, `E_2 ltW and E_1 gtW`
Photoelectric emission is possible with `lambda_1` only`
Photoelectrons experience magnetic force and move along a circular path. The galvanometer
will indicate zero deflection if the photoelectrons just complete just complete semicircular path befor
reaching the plate P. Thus `d=r=5 cm`
`:. r = 5cm = 0.05 m`
Further, r =(mv)/(Bq) = (sqrt2 Km)/(Bq)`
`:. B_(min) = (sqrt2 km)/(rq)`
Here, `K = E_1 - W = (3.1 - 2.39)`
`=0.71 eV`
Substituting the values , we have `B_(mi) = (sqrt 2xx0.71xx1.6xx10^(-`19)xx9.109xx10^(-31))/((0.05xx(1.6xx10^(-19)`
`= 5.68xx10^(-5) T`