Question

In photo electric effect cell, the cathode has work function 1.2 eV & the anode is kept at +5V with respect to the cathode.
If light of wavelength 2000 Å in incident on the cathode , find the maximum & minimum KE with which electrons.
reach the anode.
​

Answer 1

Answer:

Photons of energy 5 eV are incident on cathode. Electrons reaching the anode have kinetic energies varying from 6eV to 8eV.



This question has multiple correct options

A .

Work function of the metal is 1 eV

B .

Work function of the metal is 2 eV

C .

Current in the circuit is equal to saturation value

D .

Current in the circuit is less than saturation value.

December 30, 2019Arman Nagar

SHARE

ANSWER

According to Einstein's equation KE = hv - hv_0KE=hv−hv0

So, KE_{max}=(5-\phi )eVKEmax=(5−ϕ)eV

When these electrons are accelerated through 5V, they will reach the anode with maximum energy =(5-\phi+5)eV=(5−ϕ+5)eV

\therefore10-\phi=8∴10−ϕ=8

\phi=2eVϕ=2eV

Current is less than saturation current because if slowest electron also reached the plate it would have 5eV energy at the anode, but there it is given that the minimum energy is 6eV.

So, the answers are option (B) & (D).

Explanation:

THIS WILL HELP YOU

IF HELPFUL

THEN MARK MY ANS AS BRAINLIST