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  • A proton and an alpha are accelerated in a field of same potential energy. Then, the ratio of the Broglie wavelengths associated with the moving particles are -O

A proton and an alpha are accelerated in a field of same potential energy. Then, the ratio of the Broglie wavelengths associated with the moving particles are -

Option: 1

\mathrm{2\sqrt{2}:1}


Option: 2

\mathrm{1:2}


Option: 3

\mathrm{4:1}


Option: 4

\mathrm{2:1}


Answers (1)

best_answer

Broglie. wave length of a proton of mass \mathrm{\left(m_1\right)} and kinetic energy ( \mathrm{k} ) is given by -
\mathrm{\begin{aligned} \lambda_1 & =\frac{h}{\sqrt{2 m_1 k}} \\ \therefore \quad \lambda_1 & =\frac{h}{\sqrt{2 m_1 q v}}.......\left ( 1 \right ) \end{aligned}}
for alpha particle having mass \mathrm{m_2} carrying charge \mathrm{q_0} is acculated through potential \mathrm{v} then -
\mathrm{\lambda_2=\frac{h}{\sqrt{2 m_2 q_0 v}}}
\mathrm{\text { for } \alpha-p \text { article, } m_2=4 m \text { and } z_0=2 q}
\mathrm{\therefore \lambda_2=\frac{h}{\sqrt{2 \times 4 m, 2 q \times v}}......... \text { (2) }}
\mathrm{\therefore } Required ration
\mathrm{ \frac{\lambda_1}{\lambda_2}=\frac{4}{\sqrt{2}}=\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} }
\mathrm{ \begin{gathered} \frac{\lambda _1}{\lambda_2}=\frac{4 \sqrt{2}}{2}=\frac{2 \sqrt{2}}{1} \\ \lambda_1: \lambda_2=2 \sqrt{2}: 1 \end{gathered} }
 

Posted by

Ritika Jonwal

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