A photoelectric current emerging in the circuit of a vacuum photocell when its zinc electrode is subjected to electromagnetic radiation of wavelength 262 nm is cancelled if an external decelerating voltage 1.5 V is applied. Find the magnitude of outer contact potential difference of the given photocell. Work function of zinc metal is 3.74 eV.

Question

A photoelectric current emerging in the circuit of a vacuum photocell when its zinc electrode is subjected to electromagnetic radiation of wavelength 262 nm is cancelled if an external decelerating voltage 1.5 V is applied. Find the magnitude of outer contact potential difference of the given photocell. Work function of zinc metal is 3.74 eV. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

φ=3.74 eV λ=262 nm energy of photon =(1242 n meV /262 nm )=4.74 eV K max = eV s = hv -φ eV s =4.74-3.74=1 eV stopping potential =1 volt decelerating voltage - V B =1.5 V -1 V =0.5 V ⇒ outer contact P.D. =0.5 V

ϕ=3.74eVλ=262nm energy of photon =262nm1242nmeV​=4.74eV Kmax​=eVs​=hv−ϕ eVs​=4.74−3.74=1eV stopping potential =1 volt decelerating voltage −VB​=1.5V−1V =0.5V ⇒ outer contact P.D. =0.5V

$ϕ = 3.74 e Vλ = 262 nm$
energy of photon $= \frac{1242 n m e V}{262 nm} = 4.74 e V$
$K_{m a x} = e V_{s} = h v - ϕ$
$e V_{s} = 4.74 - 3.74 = 1 e V$
stopping potential $= 1$ volt
decelerating voltage $- V_{B} = 1.5 V - 1 V$
$= 0.5 V$
$\Rightarrow$ outer contact P.D. $= 0.5 V$