Download the app

Questions  

In a photoelectric effect experiment a metallic surface having work function 2.2eV is illuminated with a light of wavelength  4000Ao.  An incident photon makes a collision with a free electron that lies well inside the target metal and the electron makes one collision before being ejected from the surface. If 10% of the energy of the electron is lost in the collision, The kinetic energy of the ejected electron is 

a
0.28ev
b
0.90ev
c
0.86ev
d
0.59ev

detailed solution

Correct option is D

Energy of photon =hcλ=124204000ev⇒hv=3.1evEnergy lost in the collision =10100 x 3.105 ev=0.31 evNow, hv=ϕ+k+Lost energy⇒ 3.1 = 2.2 + KE + 0.31 ⇒KE=0.59 ev

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

Light described at a place by equation E=200vmsin2×1015s1t+sin10×1015s1t   falls on metal surface having work function 2 eV, the maximum kinetic energy of photoelectrons?


phone icon
whats app icon