Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of ` 0. 005 nm`.

Explanation

Atomic weight of hydrogen is 1.008 amu.
`therefore` weight of 1 mole of proton = 1.008g
`therefore` weight of 1 proton `= (1.008)/(6.022 xx 10^(23))=0.167 xx 10^(-23)g`
`=0.167 xx 10^(-26)kg`
We have, `v= (h)/(m lamda)= ((6.63 xx 10^(-34)J.s))/((0.167 xx 10^(-26)kg) (0.005 xx 10^(-9) m))= 7.94 xx 10^(4) m//s`
Kinetic energy `=(1)/(2) mv^(2) = (1)/(2) (0.167 xx 10^(-26)) (7.94 xx 10^(4))^(2)`
`=5.26 xx 10^(-18)J`
`=(5.26 xx 10^(-18))/(1.602 xx 10^(-19))eV`
= 32.8eV
As the magnitude of charge of a proton is the same as that of an electron, the potential required is equal in magnitude to the number of eV, i.e., 32.8 volts.