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What is the de - Broglie wavelength of the alpha - particle accelerated through a potential difference `V`?A. `(0.287)/(sqrt(V)) Å`B. `(12.27)/(sqrt(V)) Å`C. `(0.101)/(sqrtV)Å`D. `(0.202)/(sqrt(V))Å`

Answer» Correct Answer - C
`lambda = (h)/(sqrt( 2 m E)) = (h)/(sqrt(2 m_(alpha) Q_(alpha) V))`
On putting `Q_(alpha ) = 2 xx 1.6 xx 10^(-19)`
`m_(alpha) = 4m_(p) = 4 xx 1.67 xx 10^(-27) kg rArr lambda = (0.101)/(sqrt(V)) Å`


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