The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength

Question:

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491 \mathrm{~nm}$ is $0.710 \mathrm{~V}$. When the incident wavelength is changed to a new value, the stopping potential is $1.43 \mathrm{~V}$. The new wavelength is:

  1. (1) $400 \mathrm{~nm}$

  2. (2) $382 \mathrm{~nm}$

  3. (3) $309 \mathrm{~nm}$

  4. (4) $329 \mathrm{~nm}$


Correct Option: , 2

Solution:

(2)

From the photoelectric effect equation

$\frac{h c}{\lambda}=\phi+\operatorname{ev}_{s}$

so $\mathrm{ev}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-\phi$........

$\mathrm{ev}_{83}=\frac{\mathrm{hc}}{1}-\phi \ldots . . .(\mathrm{ii})$

Subtract equation (i) from equation (ii) ev $_{s_{1}}-\mathrm{ev}_{s_{2}}=\frac{h c}{\lambda_{1}}-\frac{h c}{\lambda_{2}}$

$v_{s_{1}}-v_{s_{2}}=\frac{h c}{e}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$

$(0.710-1.43)=1240\left(\frac{1}{491}-\frac{1}{\lambda_{2}}\right)$

$\frac{-0.72}{1240}=\frac{1}{491}-\frac{1}{\lambda_{2}}$

$\frac{1}{\lambda_{2}}=\frac{1}{491}+\frac{0.72}{1240}$

$\frac{1}{\lambda_{2}}=0.00203+0.00058$

$\frac{1}{\lambda_{2}}=0.00261$

$\lambda_{2}=383.14$

$\lambda_{2} \simeq 382 \mathrm{~nm}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now