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

• A. $1:2$
• B. $2:1$
• C. $1:\sqrt{2}$
• D. $\sqrt{2}:1$

Answer 1

Answer: (D)

Given,
$\frac{E_{A}}{E_{B}}=\frac{1}{2},\frac{U_{A}}{U_{B}}=\frac{1}{2}$
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 $K_{A}=E_{A}-U_{A}=(x-y) \ldots . .$ (i)
$K_{B}=E_{B}-U_{B}=2(x-y) \ldots .$ (ii)
$\because$ de-Broglie wavelength,
$\lambda=\frac{h}{\sqrt{2 m k}}$
So $\lambda_{A}=\frac{h}{\sqrt{2 m K_{A}}}, \lambda_{B}=\frac{h}{\sqrt{2 m K_{B}}}$
$\therefore \quad \frac{\lambda_{A}}{\lambda_{B}}=\sqrt{\frac{K_{B}}{K_{A}}} \ldots \ldots$ (iii)
From Equation (i), (ii) and (iii),
$\frac{\lambda_{A}}{\lambda_{B}}=\sqrt{\frac{2(x-y)}{(x-y)}}=\frac{\sqrt{2}}{1}$