• # question_answer The kinetic energy of electron and proton is ${{10}^{-32}}J$. Then the relation between their de-Broglie wavelengths is A) ${{\lambda }_{p}}<{{\lambda }_{e}}$ B) ${{\lambda }_{p}}>{{\lambda }_{e}}$ C) ${{\lambda }_{p}}={{\lambda }_{e}}$ D) ${{\lambda }_{p}}=2{{\lambda }_{e}}$

Correct Answer: A

Solution :

 [a] By using $\lambda =\frac{h}{\sqrt{2mE}}$ $E={{10}^{-32}}J=\text{Constante}$for both particles. Hence $\lambda \propto \frac{1}{\sqrt{m}}$ Since ${{m}_{p}}>{{m}_{e}}$ so ${{\lambda }_{p}}<{{\lambda }_{e}}.$

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