Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut - off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

Options

A. 2.16 eV
B. 216 eV
C. 21.6 eV
D. 0.216 eV

Answer

A

Explanation

Given, light of wavelength, `lambda = 488nm` and stopping potential, of photoelectrons, V = 0.38 V
As, energy of photon
`E = (hc)/(lambda) = (1250)/(488) eV`
So, `E = 2.56 eV`
`because` Maximum kinetic energy of a photo-electrons `KE_("max") = eV_(0)`
Where, `V_(0)` = stopping potential
`:. KE_("max") = eV_(0) = 0.38 eV`
From Einstein.s photoelectric equation,
`E = KE_("max") + W_(0)`
`rArr W_(0) = E - KE_("max")`
`= 2.56 - 0.38 = 2.18 ~~ 2.16 eV`
Hence, the work function of cathode material is 2.16 eV.
So, the correct option is (a).