A proton and a Li^{3+} nucleus are accelerate | vedclass

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Question

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton $(P)$ and $Li^{3+}$ nucleus accelerated by the same potential

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton $(P)$ and $Li^{3+}$ nucleus accelerated by the same potential $V$:$\frac{\lambda_{Li}}{\lambda_{P}} = \sqrt{\frac{m_{P} 	imes q_{P}}{m_{Li} 	imes q_{Li}}} = \sqrt{\frac{m_{P} 	imes e}{8.3 m_{P} 	imes 3e}} = \sqrt{\frac{1}{8.3 	imes 3}} = \sqrt{\frac{1}{24.9}} \approx \sqrt{0.04016} \approx 0.2004$.Given $\frac{\lambda_{Li}}{\lambda_{P}} = x 	imes 10^{-1}$,we have $0.2004 = x 	imes 10^{-1}$,so $x \approx 2.004$.Rounding to the nearest integer,$x = 2$.

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