Dual Nature Of Light Question 281
Question: When light of wavelength 300 nm (nanometer) falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitter [CBSE PMT 1993; JIPMER 2000]
Options:
A) 1 : 2
B) 2 : 1
C) 4 : 1
D) 1 : 4
Show Answer
Answer:
Correct Answer: B
Solution:
Work function $ =\frac{hc}{{\lambda_{0}}} $ ; where $ {\lambda_{0}} $ is threshold wavelength.
$ \therefore $ $ \frac{{W_{0_{1}}}}{{W_{0_{2}}}}=\frac{{\lambda_{0_{2}}}}{{\lambda_{0}}_{1}}=\frac{2}{1} $
BETA
