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Light of wavelength 0.6 \mu \mathrm{m} from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 \mathrm{~V}. With light of wavelength 0.4 \mu \mathrm{m} from a sodium lamp, stopping potential is 1.5 \mathrm{~V}. With this data, the value of \mathrm{h} / \mathrm{e} is:
 

Option: 1

4 \times 10^{-19} \mathrm{Vs}


Option: 2

0.25 \times 10^{15} \mathrm{Vs}


Option: 3

4 \times 10^{-15} \mathrm{Vs}


Option: 4

4 \times 10^{-8} \mathrm{VsV}


Answers (1)

best_answer

\begin{aligned} &\mathrm{As}, \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}\\ \therefore \quad \mathrm{V} & =\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \cdot \frac{\mathrm{c}}{\lambda}-\frac{\mathrm{W}}{\mathrm{e}} \\ \mathrm{V}_1 & =\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \frac{\mathrm{c}}{\lambda_1}-\frac{\mathrm{W}}{\mathrm{e}} \\ \mathrm{V}_2 & =\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \frac{\mathrm{c}}{\lambda_2}-\frac{\mathrm{W}}{\mathrm{e}} \end{aligned}
Solving these two equations, we get
\frac{\mathrm{h}}{\mathrm{e}}=\frac{\lambda_1 \lambda_2\left(\mathrm{~V}_1-\mathrm{V}_2\right)}{\mathrm{c}\left(\lambda_2-\lambda_1\right)}=\frac{\left(0.6 \times 0.4 \times 10^{-12}\right)(1.0)}{\left(3 \times 10^8\right)\left(0.2 \times 10^{-6}\right)}=4 \times 10^{-15} \mathrm{Vs}
 

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