Question

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:
(1) 0.65 eV
(2) 1.0 eV
(3) 1.3 eV
(4) 1.5 eV

​Answer is

The kinetic energy of the photoelectrons emitted from the metal surface is given by: E =hν−ϕ
Here, hν is the energy of the incident radiation and is the work function of the metal. Initially, we have:0. 5 =hν−ϕ ...(i) When the energy of the incident radiation is increased by 20%, we have:0. 8 = 1.2hν−ϕ ...(ii)
Solving (i) and (ii), we get:ϕ = 1.0 eV
Hence, the correct option is (2).
Of what value 20% is taken?

Please answer as soon as possible....

Answer 1

the 20% value should be taken of hν as it is the energy of the radiation used

Answer 2

Answer:

20% is the energy increased for the change in kinetic energy to 0.8 eV