The de Broglie wavelength of a particle accel | vedclass

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.This implies that $\lambda \propto \fr

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.This implies that $\lambda \propto \frac{1}{\sqrt{V}}$.Given: $\lambda_1 = 10^{-10} \ m = 1 \ \mathring A$ at $V_1 = 150 \ V$.We need to find $\lambda_2$ at $V_2 = 600 \ V$.Using the ratio: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.Substituting the values: $\frac{1}{\lambda_2} = \sqrt{\frac{600}{150}} = \sqrt{4} = 2$.Therefore,$\lambda_2 = \frac{1}{2} = 0.5 \ \mathring A$.

The de Broglie wavelength of a particle accelerated through a potential difference of $150 \ V$ is $10^{-10} \ m$. If it is accelerated through a potential difference of $600 \ V$,what will be its wavelength in $\mathring A$?

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