What is the de - Broglie wavelength of the alpha - particle accelerated through a potential difference `V`?

Options

A. `12.27/(sqrt(V)) Å`
B. `0.202/(sqrt(V)) Å`
C. `0.101/(sqrt(V)) Å`
D. `2.87/(sqrt(V)) Å`

Answer

c

Explanation

De-broglie wavelength, `lambda=h/(sqrt(2mqV))`
`:. lambda=(6.6xx10^(-34))/(sqrt(2xx(4xx1.66xx10^(-27))xx2xx1.6xx10^(-19)V))`
`=(0.101xx10^(-10))/(sqrt(V))m=0.101/(sqrt(V))Å`