Question

When light of wavelength $ 300\,nm $ (nanometer) falls on a photoelectric emitter, photoelectrons are just liberated. For another emitter, however, light of $ 600\,nm $ wavelength is sufficient for creating photoemission. What is the ratio of the work function of the two emitters?

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

Answer 1

Answer: (B)

The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal.
Given by $W=h v \ldots$ (i)
where $v$ is threshold frequency and $h$ is Planck's constant.
Also, $v=\frac{c}{\lambda}$ ...(ii)
where $c$ is speed of light and $X$ the threshold wavelength.
From Eqs. (i) and (ii), we get
$W=\frac{h c}{\lambda}$
Given, $\lambda_{1}=300 \,nm , \,\,\,\,\lambda_{2}=600\, nm $
$\therefore \frac{\lambda_{2}}{\lambda_{1}}=\frac{W_{1}}{W_{2}} $
$\Rightarrow \frac{W_{1}}{W_{2}}=\frac{600}{300}=\frac{2}{1}$