Two electrons in two hydrogen-like atoms $A$ and $B$ have their total energies $E_A$ and $E_B$ in th

Question

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 $\lambda_A$ and $\lambda_B$ are their de-Broglie wavelengths, then $\lambda_A: \lambda_B$ is:

Professor PC Thomas Classes · Answer

$$
\frac{E_A}{E_B}=\frac{1}{2}, \frac{U_A}{U_B}=\frac{1}{2}
$$So $E_A=x, E_B=2 x$
And $U_A=y, U_B=2 y$
$$
\because E_A=U_A+K_A
$$And $E_B=U_B+K_B$
here $K_A$ and $K_B$ are kinetic energy of particles $A$ and $B$
So $\boldsymbol{K}_{\boldsymbol{A}}=\boldsymbol{E}_{\boldsymbol{A}}-\boldsymbol{U}_{\boldsymbol{A}}=(\boldsymbol{x}-\boldsymbol{y})$
$$
K_B=E_B-U_B=2(x-y)
$$∵ de-Broglie wavelength, $$
\lambda=\frac{h}{\sqrt{2 m k}}
$$So $\lambda_A=\frac{h}{\sqrt{2 m K_A}}, \lambda_B=\frac{h}{\sqrt{2 m K_B}}$
$$
	herefore \quad \frac{\lambda_A}{\lambda_B}=\sqrt{\frac{K_B}{K_A}}
$$From Equation (i), (ii) and (iii),
$$
\frac{\lambda_A}{\lambda_B}=\sqrt{\frac{2(x-y)}{(x-y)}}=\frac{\sqrt{2}}{1}
$$