Dual Nature of Radiation and Matter - Result Question 66
70. 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 emitters?
[1993]
(a) $1: 2$
(b) $2: 1$
(c) $4: 1$
(d) $1: 4$
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Answer:
Correct Answer: 70. (b)
Solution:
(b) $W_0=\frac{h c}{\lambda_0}$ or $W_0 \propto \frac{1}{\lambda_0}$;
$\Rightarrow \frac{W_1}{W_2}=\frac{\lambda_2}{\lambda_1}=\frac{600}{300}=2$
