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  • In a photo emissive cell with executing wavelength  , the fastest electron has speed <img alt="v" src="http

In a photo emissive cell with executing wavelength \lambda , the fastest electron has speed v. If the exciting wavelength is changed to 3\lambda /4 , the speed of the fastest emitted electron will be -

Option: 1

v\left ( 3/4 \right )^{1/2}


Option: 2

v\left ( 4/3 \right )^{1/2}


Option: 3

Less than v\left ( 4/3 \right )^{1/2}


Option: 4

Greater than v\left ( 4/3 \right )^{1/2}


Answers (1)

best_answer

\begin{aligned} & \mathrm{v}_{\max }=\sqrt{\frac{3 \mathrm{~h}}{\mathrm{~m}}\left(\frac{\mathrm{hc}}{\lambda}-\mathrm{W}\right)} \\ & \lambda_2<\lambda_1 \\ & \mathrm{v}_{\text {max }_2}>\mathrm{v}_{\text {max }_1} \end{aligned}
\mathrm{v}_{\max _2}=\sqrt{\frac{3 \mathrm{~h}}{\mathrm{~m}}\left(\frac{\mathrm{hc}}{\frac{3 \lambda}{4}}-\mathrm{W}\right)}=\sqrt{\frac{4}{3}\left[\frac{3 \mathrm{~h}}{\mathrm{~m}}\left(\frac{\mathrm{hc}}{\lambda}-\frac{3}{4} \mathrm{~W}\right)\right]}
\mathrm{v}_{\max _2}>\sqrt{\frac{4}{3}} \mathrm{v}_{\max _1}

Posted by

sudhir kumar

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