Question

An alpha partcle and a proton are accaelerated from rest by the same potential.Find the ratio of their de broglie wavelengths

Answer 1

we have formula

de broglie wavelength = h/√2MeV

where M= mass of the particle

e= charge of the particle

V= accelerating potential

now an alpha particle has 4 times mass as that of proton and two times as that of charge of proton

so let de broglie wavelength of proton be d1=h/√2MeV    -----1

so for alpha particle d2=h/√2*4M*2e*V  ----2

same accelerating potential given so V remains same for both

1/2

d1/d2=(h/√2*M*e*V)*(√2*4M*2e*V/h)

d1/d2=(√16*M*e*v)/(√2*M*e*V)

d1/d2= 4/√2

d1/d2=(2 * √2 *√2)/√2

d1/d2=2√2/1

hence this is the required ratio of d1 and d2 that is 2√2 is to 1

hope this helps