Two particles A and B of same mass have their de-Broglie wavelengths in the ratio X A : X B = K : 1 . Their potential energies U A : U B = 1 : K 2 . The ratio of their total energies E A : E B is

Step-by-step solution

According to question, X A X B = K 1 ...(i) U A U B = 1 K 2 So, U A = y and U B = K 2 y According to de-Broglie wavelength, λ = h 2 m K 1 .....(ii) Here, K 1 is kinetic energy. So, from Equations (i) and (ii), ⇒ h 2 m K 1 A h 2 m K 1 B = K 1 ⇒ K 1 B K 1 A = K 1 ⇒ K 1 B K 1 A = K 2 So, K 1 B = K 2 x and K 1 A = x ∵ Total energy, E = K 1 + U So, E A = K 1 A + U A and E B = K 1 B + U B or E A E B = K 1 A + U A K 1 B + U B = x + y K 2 x + K 2 y ⇒ So, E A E B = 1 K 2 o r E A : E B = 1 : K 2