Question

When light of wavelength $300$$nm$ falls on a photoelectric emitter, photo-electrons are liberated. For another emitter, light of wavelength $600$$nm$ is sufficient for liberating photo-electrons. The ratio of the work function of the two emitters is:

• A. $1:2$
• B. $4:1$
• C. $2:1$
• D. $1:4$

Answer 1

Answer: (C)

Minimum energy required to emit photoelectrons from an emitter, is its work potential.

Thus , $W_{0_1}=\frac{hc}{{\lambda }_1}$

$W_{0_2}=\frac{hc}{{\lambda }_2}$

$⟹\frac{W_{0_1}}{W_{0_2}}=\frac{{\lambda }_2}{{\lambda }_1}$

$=\frac{600}{300}=2:1$

Hence correct answer is option B.