Calculate the accelerating potential that must be applied on a proton beam to give it an effective wavelength of 0.005 nm ?

Calculate the accelerating potential that must be applied on a proton

1.	Calculate the accelerating potential that must be applied on a proton beam to give it an effective wavelength of 0.005 nm ?
Answer» SOLUTION :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`