Question

An electron is in an excited state in a hydrogen like atom. It has a total energy of $0.34eV$. The kinetic energy of the electron is $E$ and its de-Bronglie wavelength is $\lambda$. Then: (Take $m_e=9.1\times 10^{-31}kg)$

• 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

Answer: (B)

$P.E=-2\times K.E$
$=-2E$
$\therefore$ Total energy $=-2E+E=-E=-3.4eV$
$\therefore E=3.4eV$
$\Rightarrow E=3.4\times 1.6\times 10^{-19}J$
Let $P=$ momentum and $m=$ mass of the electron
$\therefore E=\frac{P^2}{2m}$
or $P=\sqrt{2mE}$
$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2mE}}$
$=\frac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 0.34\times 1.6\times 10^{-19}}}$
$\lambda =6.6\times 10^{-10}m$