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(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_p qV}}$.Substituting the values: $h = 6.63 	imes 10^{-

Vedclass · Accepted Answer

$9.1 	imes 10^{-13} \,m$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_p qV}}$.Substituting the values: $h = 6.63 	imes 10^{-34} \,J-s$, $m_p = 1.67 	imes 10^{-27} \,kg$, $q = 1.6 	imes 10^{-19} \,C$, and $V = 1000 \,V$.$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 1.67 	imes 10^{-27} 	imes 1.6 	imes 10^{-19} 	imes 1000}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{5.344 	imes 10^{-43}}} = \frac{6.63 	imes 10^{-34}}{7.31 	imes 10^{-22}} \approx 9.07 	imes 10^{-13} \,m$.Thus, the wavelength is approximately $9.1 	imes 10^{-13} \,m$.

If a proton is accelerated through a potential difference of $1000 \,V$, then its de-Broglie wavelength is (given, $m_p = 1.67 \times 10^{-27} \,kg$, $h = 6.63 \times 10^{-34} \,J-s$).

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