Question

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

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

Answer 1

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

The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.
$PE=2E$
The total energy is: $TE=PE+KE$
$-3.4=-2\times 3.4+KE$
$KE=3.4eV$
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:
$\lambda =\frac{h}{p}$
$\lambda =\frac{h}{\sqrt{2}mE}$
$\lambda =\frac{6.6\times {10}^{34}}{\sqrt{2}\times 9.1\times {10}^{-31}\times (-3.4)\times 1.6\times {10}^{-19}}$

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

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

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