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  • The ratio of wavelengths of proton and deuteron accelerated by potential <img alt="\mathrm{V_{p}\: and\: \: \mathrm{V}_{\mathrm{d}}}" src="/latex-image/?%5Cmathrm%7BV_%7Bp%7D%5C%3A%20and%5C%3A%20%5

The ratio of wavelengths of proton and deuteron accelerated by potential \mathrm{V_{p}\: and\: \: \mathrm{V}_{\mathrm{d}}} is 1: \sqrt{2}. Then, the ratio of \mathrm{V}_{\mathrm{p}}$ to $\mathrm{V}_{\mathrm{d}} will be :

Option: 1

1:1


Option: 2

\sqrt{2}:1


Option: 3

2:1


Option: 4

4:1


Answers (1)

best_answer

\mathrm{\begin{gathered} \lambda_p=\frac{h}{p}=\frac{h}{m_d v_p}=\frac{h}{\sqrt{2 m_p q_p V_p}} \\ \end{gathered}}

\mathrm{\lambda_d=\frac{h}{p}=\frac{h}{m_d v_d}=\frac{h_1}{\sqrt{2 m_d q_d V_d}} }

\mathrm{\frac{\lambda_p}{\lambda_d}=\frac{1}{\sqrt{2}}=\frac{\sqrt{m_d} q_d V_d}{\sqrt{m_p q_p V_p}} }

\mathrm{\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{1} \times \frac{\sqrt{V_d}}{\sqrt{v_p}} }

\mathrm{\frac{V_p}{V_d}=\frac{4}{1}}

Hence (4) is correct option




 

Posted by

Shailly goel

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