Question

An electron is in an excited state in a hydrogen like atom. It has a total energy of $-3.4$ $eV$. The kinetic energy of the electron is $E$ and its de Broglie wavelength is $\lambda$.

• A. $E=6.8$ $eV$, $\lambda =-6.6\times {10}^{-10}$ $m$
• B. $E=3.4$ $eV$, $\lambda =-6.6\times {10}^{-10}$ $m$
• C. $E=3.4$ $eV$, $\lambda =-6.6\times {10}^{-11}$ $m$
• D. $E=6.8$ $eV$, $\lambda =-6.6\times {10}^{-11}$ $m$

Answer 1

The correct option is B $E=3.4$ $eV$, $\lambda =-6.6\times {10}^{-10}$ $m$

The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.
$PE=-2E$

The total energy is:
$TE=PE+KE$
$-3.4=-2E+E$
$E=3.4eV$

Let, p be the momentum of electron and m be the mass of electron.
$E=\frac{p^2}{2m}$
$p=\sqrt{2}mE$

Now, the De-Broglie wavelength associated with an electron is
$\lambda =\frac{h}{p}$
$\lambda =\frac{h}{\sqrt{2}mE}$
$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2}\times 9.1\times {10}^{-31}\times (-3.4)\times 1.6\times {10}^{-19}}$

$\lambda =-\frac{6.6\times {10}^{-34}}{9.95\times {10}^{-25}}$

$\lambda =-0.66\times {10}^{-9}$

$\lambda =-6.6\times {10}^{-10}m$