Question

A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which light energy of 1.0 × 10⁻⁷ J falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball?

(Use h = 6.63 × 10^-34J-s = 4.14 × 10^-15 eV-s, c = 3 × 10⁸ m/s and me = 9.1 × 10^-31kg)

Answer 1

Given:-

Radius of the silver ball, r = 4.8 cm

Wavelength of the ultra violet light, λ = 200 nm = 2 × 10⁻⁷ m

Total energy of light, E = 1.0 × 10⁻⁷ J

We are given that one photon out of every ten thousand is able to eject a photoelectron.

Energy of one photon,

`E^' = (hc)/lambda`,

where h = Planck's constant  
            c = speed of light
            `lambda` = wavelength of light

On substituting the respective values in the above formula, we get :

`E^' = (6.63 xx 10^-34 xx 3 xx 10^8)/(2 xx 10^-7)`

`=9.945 xx 10^-19`

Number of photons,

`n = E/E^' = (1 xx 10^-7)/(9.945 xx 10^-19) = 1 xx 10^11`

Number of photoelectrons

= `(1 xx 10^11)/10^4 = 1 xx 10^7`

The amount of positive charge developed due to the outgoing electrons,

`q = 1 xx 10^7 xx 1.6 xx 10^-19`

`= 1.6 xx 10^-12 C`

Potential developed at the centre as well as on surface,

`V = (Kq)/r`,

where K = `1/(4piε_0)`

`therefore V = (9 xx 10^9 xx 1.6 xx 10^-12)/(4.8 xx 10^-2) = 0.3 V`

Potential inside the silver ball remains constant. Therefore, potential at the centre of the sphere is 0.3 V.