Question

Starting from rest, an electron accelerates through a potential difference of 40 kV. (a) What is its de Broglie wavelength? (b) What is the highest frequency of the photon produced by this electron when it is stopped by hitting a target?

Answer 1

The given start from rest, electron accelerate through a p.d of 40Kv

the de broglie wavelength w,

for a particle with momentum p,

is given by the formula,

w=h/p,

h is plank's constant

so we find the p

we calculate the energy of the electron E

w=Vxq

w is the energy transfer

charge q is moved through a pd V

assuming that this is done in a vaccum and that the electron has an initial KE of zero

then the W=E=KE100ev

now the another equation that links p with E

which E is p^2/2m m is mass of electron

rearrage given p-(2mE)^1/2

so

p=(2*9.1*10^-31*1.6*10^-17)^1/2

therefore

p=5.39*10^-24kg m/s

w=h/p=1.23*10^-10m

or

w=0.123 nm,