Question

A proton and electron har de broglie wavelength of 1.6nm
Find the ratio of their momenta.
Compare the ke of proton with that of electron​

Answer 1

Explanation:

λ=h/p

where,h is constant

p = momentum

λ1/λ2=p2/p1

h cancel out each other

given λ1=λ2=1.6nm

1.6nm=p2/p1

p1/p2=1/1.6nm

=0.62×10^9

=6.2×10^8

k.E=p^2/2m

p=h/λ

K.Ee/K.Ep=(h/λ)^2 e/2m e ×2mp/(h/λ)^2 p

k.E e/k.E p=mp/me