Dual Nature Of Light Question 437
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) $ .
Options:
1.6
6
2
1.2
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ 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
