The de-Broglie wavelength of a particle accel | vedclass

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(B) The de-Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Sin

Vedclass · Accepted Answer

$0.5$

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(B) The de-Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Since $\lambda \propto \frac{1}{\sqrt{V}}$,we can write the ratio as:$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.Given $\lambda_1 = 10^{-10} \ m = 1 \ \mathring{A}$,$V_1 = 150 \ V$,and $V_2 = 600 \ V$.Substituting the values:$\frac{1 \ \mathring{A}}{\lambda_2} = \sqrt{\frac{600}{150}} = \sqrt{4} = 2$.Therefore,$\lambda_2 = \frac{1}{2} \ \mathring{A} = 0.5 \ \mathring{A}$.

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

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