Question

Two particles A and B of same mass have their de Broglie wavelengths in the ratio k:1 . Their potential energies U:U' = 1:k . The ratio of their total energies is

Answer 1

Total energies , T.E = K.E + P.E
K.E denotes kinetic energy and P.E denotes Potential energy .
∵ kinetic energy , K.E = P²/2m , here P is momentum and m is mass of body .
∴ K.E $\propto$ P²

According to de - broglie's wavelength ,
λ = h/P ,h is plank's constant and λ is wavelength
∴ K.E $\propto$ 1/λ² -------(1)

Now, ratio of kinetic energy of bodies = 1 : k² [ from equation (1) and ratio of wavelength = k : 1 ]
e.g., K.E₁/K.E₂ = 1/k²
K.E₂= k² K.E₁ -------(2)
Given, P.E₁/P.E₂ = 1 : k²[you did mistake in typing , ratio is 1 : k²]
P.E₂ = k²P.E₁ ----------------(3)
∴ ratio of total energies = ratio of (K.E + P.E)
= (K.E₁ + P.E₁)/(K.E₂ + P.E₂)
= (K.E₁ + P.E₁)/(k² K.E₁ + k² P.E₁)
= (K.E₁ + P.E₁)/(k²)(K.E₁ + P.E₁)
= 1/k²

Hence, ratio of total energies = 1 : k²