A particle having electric charge $3 imes 10^{-19} \mathrm{C}$ and mass $6 imes 10^{-27} \mathrm

Question

A particle having electric charge $3 	imes 10^{-19} \mathrm{C}$ and mass $6 	imes 10^{-27} \mathrm{~kg}$ is accelerated by applying an electric potential of 1.21 V . Wavelength of the matter wave associated with the particle is $\alpha 	imes 10^{-12} \mathrm{~m}$. The value of $\alpha$ is $\_\_\_\_$ . (Take Planck's constant $=6.6 	imes 10^{-34} \mathrm{~J} . \mathrm{s}$ )

Professor PC Thomas Classes · Answer

The kinetic energy of the accelerated particle is $K = qV$. The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$. Substitute the given values: $2mqV = 2 	imes (6 	imes 10^{-27}) 	imes (3 	imes 10^{-19}) 	imes 1.21 = 36 	imes 10^{-46} 	imes 1.21$. Its square root is $\sqrt{2mqV} = 6 	imes 1.1 	imes 10^{-23} = 6.6 	imes 10^{-23}	ext{ kg m/s}$. The wavelength is $\lambda = \frac{6.6 	imes 10^{-34}}{6.6 	imes 10^{-23}} = 10^{-11}	ext{ m} = 10 	imes 10^{-12}	ext{ m}$. Comparing with $\alpha 	imes 10^{-12}	ext{ m}$, we get $\alpha = 10$.