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  • light of wavelength strikes a photoelectric surface and electrons are ejected with kinetic energy k. If kn

light of wavelength\lambda strikes a photoelectric surface and electrons are ejected with kinetic energy k. If k is to be increased to exactly twice its original value, the wavelength must be changed to \lambda^{\prime} such that

Option: 1

\lambda^{\prime}<\frac{\lambda}{2}


Option: 2

\lambda^{\prime}>\frac{\lambda}{2}


Option: 3

\lambda>\lambda^{\prime}> \frac{\lambda}{2}


Option: 4

\lambda^{\prime}=\frac{\lambda}{2} .


Answers (1)

best_answer

\mathrm{\begin{aligned} & \frac{h c}{\lambda}=E+W_0 \\ & \frac{h c}{\lambda^{\prime}}=2 E+W_0 \end{aligned}}

 Dividing (1) \& (2), we get 

\mathrm{\frac{\lambda^2}{\lambda}=\frac{E+\omega_0}{2 E+\omega_0} \Rightarrow \frac{\lambda^1}{\lambda}<1 \Rightarrow \lambda^{\prime}<\lambda .}

\mathrm{\text { Also } \frac{\lambda^1}{\lambda}=\frac{1}{2} \cdot\left(\frac{E+W_0}{E+\frac{W_0}{2}}\right)>\frac{1}{2} .}

\mathrm{\begin{aligned} & \Rightarrow \lambda^{\prime}>\frac{\lambda}{2} . \\ & \therefore \quad \lambda>\lambda^{\prime}>\frac{\lambda}{2} . \end{aligned}}

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shivangi.shekhar

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