1.

What is the de-Broglie wavelength of the alpha-particle accelerated through a potential difference of V volt ? (mass of alpha-particle = 6.6455 xx 10^(-27) kg)

Answer»

`(0.287)/(sqrt(V)) Å`
`(12.27)/(sqrt(V)) Å`
`(0.101)/(sqrt(V)) Å`
`(0.202)/(sqrt(V)) Å`

Solution :K = QV = 2eV
`LAMBDA = (h)/(sqrt(2mK)) = (h)/(sqrt(2M xx 2eV)) = (h)/(sqrt(4meV))`
`= (6.63 xx 10^(-34))/(sqrt(4 xx 6.6465 xx 10^(-27) xx 1.6 xx 10^(-19) V)) m = (6.63 xx 10^(-11))/(6.52 sqrt(V))m = (0.101)/(sqrt(V)) Å`


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