Two electrons in two hydrogen-like atoms A and B have their total energies E A and E B in the ratio E A : E B = 1 : 2 . Their potential energies U A and U B are in the ratio U A : U B = 1 : 2 . If λ A and λ B are their de-Broglie wavelengths, then λ A : λ B is

Step-by-step solution

Given, E A E B = 1 2 , U A U B = 1 2 So E A = x , E B = 2 x And U A = y , U B = 2 y ∵ E A = U A + K A And E B = U B + K B here K A and K B are kinetic energy of particles A and B So K A = E A - U A = x - y .....(i) K B = E B - U B = 2 x - y ....(ii) ∵ de-Broglie wavelength, λ = h 2 m k So λ A = h 2 m K A , λ B = h 2 m K B ∴ λ A λ B = K B K A .....(iii) From Equation (i), (ii) and (iii), λ A λ B = 2 x - y x - y = 2 1