Dual Nature Of Light Question 341
Question: A photon of energy 8 eV is incident on metal surface of threshold frequency $ 1.6\times 10^{15}Hz. $ The maximum kinetic energy of the photoelectrons emitted (in eV) (Take $ h=6\times {{10}^{-34}}Js) $ . [MP. PET 2005]
Options:
A) 1.6
B) 6
C) 2
D) 1.2
Show Answer
Answer:
Correct Answer: C
Solution:
$ K.E.=h\nu -h{\nu_{0}} $ $ =8\ eV-( \frac{6\times {{10}^{-34}}\times 1.6\times 10^{15}}{1.6\times {{10}^{-19}}}eV ) $ = $ 8\ eV-6\ eV=2\ eV $
BETA
