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If \mathrm{E_1, E_2, E_3} are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
 

Option: 1

\mathrm{E_1>E_3>E_2}


Option: 2

\mathrm{E_2>E_3>E_1}


Option: 3

\mathrm{ E_1>E_2>E_3}


Option: 4

\mathrm{ E_1=E_2=E_3}


Answers (1)

best_answer

According to relation, \mathrm{ \mathrm{E}=\frac{1}{2} \mathrm{mv}^2}

\begin{aligned} & \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}}=\mathrm{v} \\ & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\\ &\text{Because }\mathrm{m}_1<\mathrm{m}_3<\mathrm{m}_2 \end{aligned}

So for same  \lambda,

\mathrm{\mathrm{E}_1>\mathrm{E}_3>\mathrm{E}_2.}

Posted by

Anam Khan

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