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For what kinetic energy of proton, will the associated de Broglie wavelength be 16.5 \mathrm{~nm}? (\text{Given } \mathrm{m}_{\mathrm{p}}=1.675 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.63 \times 10^{-34} \mathrm{Js} )
 

Option: 1

5.2 \times 10^{-4} \mathrm{~J}


Option: 2

5.2 \times 10^{-20} \mathrm{~J}


Option: 3

4.8 \times 10^{-25} \mathrm{~J}


Option: 4

4.8 \times 10^{-30} \mathrm{~J}


Answers (1)

best_answer

Kinetic energy of proton, \mathrm{K}=\frac{1}{2} \mathrm{mv}^2

\begin{aligned} & =\frac{1}{2 \mathrm{~m}}(\mathrm{mv})^2=\frac{1}{2 \mathrm{~m}}\left(\frac{\mathrm{h}^2}{\lambda^2}\right) \quad\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right) \\ & =\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times 1.675 \times 10^{-27} \times\left(16.5 \times 10^{-9}\right)^2}=4.82 \times 10^{-25} \mathrm{~J} \end{aligned}

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