Calculate the accelerating potential that must be imported to a proton beam to give it an effective wavelength of 0.05 nm

Explanation

Step 1. Calculation of velocity of proton
Mass of proton `~=` Mass of hydrogen atom `= (1.008)/(6.022 xx 10^(23)) g = 1.67 xx 10^(-24) g = 1.67 xx 10^(-27) kg`
Wavelength `(lamda) = 0.005nm = 0.005 xx 10^(-9) m = 5 xx 10^(-12)m`
Applying de Broglie equaiton,
`lamda = (h)/(mv) or v = (h)/(m xx lamda)= ((6.626 xx 10^(-34) kg m^(2) s^(-1)))/((1.67 xx 10^(-27) kg) (5 xx 10^(12) m)) = 7.94 xx 10^(4) m s^(-1)`
Step II. Calculate of accelerating potential
If accelerating potential is V volts, then energy acquired by the proton = eV where e is the charge on the proton (which is equal to the charge on the electron). This becomes the kinetic energy of the proton.
Hence, `e xx V = (1)/(2) mv^(2) or V = (mv^(2))/(2 xx e) = ((1.67 xx 10^(-27) kg)(7.94 xx 10^(4) ms^(-1)))/(2 xx (1.602 xx 10^(-19) C)) = 32.8 (kg m^(2) s^(-1)) (C^(-1))`
`= 32.8 J C^(-1) = 32.8 (CV) C^(-1) = 32.8 V`