Search any question & find its solution
Question:
Answered & Verified by Expert
Two light waves of wavelengths $600 \mathrm{~nm}$ and $200 \mathrm{~nm}$ incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is $\frac{1}{3}$ of the maximum velocity of the photoelectrons produced due to the other wavelength, then the work function of the metal is
Options:
Solution:
1308 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{hc}}{8} \times 10^7 \mathrm{~J}$
By Einstein photoelectric equation
$\begin{aligned} & \frac{\mathrm{hc}}{\lambda_1}=\phi+\frac{1}{2} \mathrm{mV}^2, \text { where } \lambda_1=200 \mathrm{~nm} \\ & \frac{\mathrm{hc}}{\lambda_2}=\phi+\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{V}}{3}\right)^2, \text { where } \lambda_2=600 \mathrm{~nm}\end{aligned}$
Subtracting (ii) from (i), we get
$\Rightarrow \frac{\frac{\mathrm{hc}}{\lambda_1}-\phi}{\frac{\mathrm{hc}}{\lambda_2}-\phi}=9$
$\begin{aligned} & \Rightarrow \frac{\mathrm{hc}}{\lambda_1}-\phi=\frac{9 \mathrm{hc}}{\lambda_2}-9 \phi \\ & \Rightarrow 8 \phi=\mathrm{hc}\left(\frac{9}{\lambda_2}-\frac{1}{\lambda_1}\right) \\ & \Rightarrow \phi=\frac{\mathrm{hc}}{8}\left(\frac{9}{600}-\frac{1}{200}\right) \times \frac{1}{10^{-9}} \\ & \Rightarrow \phi=\frac{\mathrm{hc}}{8}\left(\frac{9-3}{600}\right) \times \frac{1}{10^{-9}} \\ & \Rightarrow \phi=\frac{\mathrm{hc}}{8} \times 10^7 \mathrm{~J}\end{aligned}$
$\begin{aligned} & \frac{\mathrm{hc}}{\lambda_1}=\phi+\frac{1}{2} \mathrm{mV}^2, \text { where } \lambda_1=200 \mathrm{~nm} \\ & \frac{\mathrm{hc}}{\lambda_2}=\phi+\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{V}}{3}\right)^2, \text { where } \lambda_2=600 \mathrm{~nm}\end{aligned}$
Subtracting (ii) from (i), we get
$\Rightarrow \frac{\frac{\mathrm{hc}}{\lambda_1}-\phi}{\frac{\mathrm{hc}}{\lambda_2}-\phi}=9$
$\begin{aligned} & \Rightarrow \frac{\mathrm{hc}}{\lambda_1}-\phi=\frac{9 \mathrm{hc}}{\lambda_2}-9 \phi \\ & \Rightarrow 8 \phi=\mathrm{hc}\left(\frac{9}{\lambda_2}-\frac{1}{\lambda_1}\right) \\ & \Rightarrow \phi=\frac{\mathrm{hc}}{8}\left(\frac{9}{600}-\frac{1}{200}\right) \times \frac{1}{10^{-9}} \\ & \Rightarrow \phi=\frac{\mathrm{hc}}{8}\left(\frac{9-3}{600}\right) \times \frac{1}{10^{-9}} \\ & \Rightarrow \phi=\frac{\mathrm{hc}}{8} \times 10^7 \mathrm{~J}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.