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  • Light of wavelength , strikes a photoelectric surface and electrons are ejected with an energy <img alt="\mathrm{E}" src="https://entrancecorner.o

Light of wavelength \lambda, strikes a photoelectric surface and electrons are ejected with an energy \mathrm{E}. If \mathrm{E} is to be increased to exactly twice its original value, the wavelength changes to \lambda^{\prime}, where:
 

Option: 1

\lambda^{\prime} is less than \lambda / 2
 


Option: 2

\lambda^{\prime}  is greater than \lambda / 2

 


Option: 3

\lambda^{\prime}  is greater than \lambda / 2 but less than \lambda
 


Option: 4

\lambda^{\prime}  is exactly equal to \lambda / 2


Answers (1)

best_answer

Energy of photoelectron
\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi_0 \Rightarrow \frac{\mathrm{hc}}{\lambda}=\mathrm{E}+\phi_0 \quad \quad \quad(i)
Where \phi_0  is the work function for the metal surface (constant).

When E^{\prime}=2 E, then
\begin{aligned} & 2 \mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\phi_0 \\ & =\frac{\mathrm{hc}}{\lambda^{\prime}}=2 \mathrm{E}+\phi_0 \quad \quad\quad(ii) \end{aligned}
Dividing eq. (i) by eq. (ii), we get


 \begin{aligned} &\frac{\lambda^{\prime}}{\lambda}=\frac{\mathrm{E}+\phi_0}{2 \mathrm{E}+\phi_0}\\ & \frac{\lambda^{\prime}}{\lambda}=\frac{\left(\mathrm{E}+\phi_0\right)}{2\left(\mathrm{E}+\frac{\phi_0}{2}\right)} \\& \therefore \quad \frac{\lambda^{\prime}}{\lambda} >\frac{1}{2} \\ &\text { or } \quad \lambda^{\prime} >\frac{\lambda}{2} \\ & \therefore \quad \lambda>\lambda^{\prime} >\frac{\lambda}{2} \end{aligned}

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vishal kumar

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