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 $E_A=x,E_B=2x$
And $U_A=y,U_B=2y$
$\because E_A=U_A+K_A$
And $E_B=U_B+K_B$
here $K_A$ and $K_B$ are kinetic energy of particles $A$ and $B$
So $K_A=E_A-U_A=(x-y)\dots ..$ (i)
$K_B=E_B-U_B=2(x-y)\dots .$ (ii)
$\because$ de-Broglie wavelength,
$\lambda =\frac{h}{\sqrt{2mk}}$
So ${\lambda }_A=\frac{h}{\sqrt{2m{K}_A}},{\lambda }_B=\frac{h}{\sqrt{2m{K}_B}}$
$\therefore \frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{K_B}{K_A}}\dots \dots$ (iii)
From Equation (i), (ii) and (iii),
$\frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{2(x-y)}{(x-y)}}=\frac{\sqrt{2}}{1}$