The de-Broglie wavelength of a particle accel | vedclass

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(B) The de-Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the relation $\lambda = \frac

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

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