The de-Broglie wavelength of proton `(" charge" = 1.6 xx 10^(-19)C, "mass" = 1.6 xx 10^(-27) Kg)` accelerated through a potential difference of 1kV is

Options

A. ` 600Å`
B. 0.9 pm
C. ` 7Å`
D. ` 0.9Å`

Answer

B

Explanation

de Broglie wavelength ,
`lamda=h/p=h/sqrt(2mE)=h/(sqrt(2mqV))`
`:. lamda =(6.6 xx10^(-34))/(sqrt(2xx(1.6 xx10^(-27))xx(1.6 xx10^(-19)) xx1000)`
or `lamda=(6.6xx10^(-34)xx10^(22) )/(1.6 sqrt20 )= 0.9 xx10^(-12) m = 0.9 `pm