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(B) Given that the de Broglie wavelengths are equal: $\lambda_{e} = \lambda_{N}$.Using the de Broglie relation $\lambda = \frac{h}{mv}$,we have $\frac

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$1758$

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(B) Given that the de Broglie wavelengths are equal: $\lambda_{e} = \lambda_{N}$.Using the de Broglie relation $\lambda = \frac{h}{mv}$,we have $\frac{h}{m_{e} v_{e}} = \frac{h}{m_{N} v_{N}}$.This simplifies to $m_{e} v_{e} = m_{N} v_{N}$,or $v_{e} = \frac{m_{N}}{m_{e}} v_{N}$.Given $v_{e} = x \ v_{N}$,we have $x = \frac{m_{N}}{m_{e}}$.Substituting the values: $x = \frac{1.6 	imes 10^{-27} \ kg}{9.1 	imes 10^{-31} \ kg}$.$x = \frac{1.6}{9.1} 	imes 10^{4} = 0.175824 	imes 10^{4} = 1758.24$.The nearest integer value of $x$ is $1758$.

The wavelength of an electron and a neutron will become equal when the velocity of the electron is $x$ times the velocity of the neutron. The value of $x$ is (Nearest Integer). (Mass of electron is $9.1 \times 10^{-31} \ kg$ and mass of neutron is $1.6 \times 10^{-27} \ kg$)

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