Question

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

• 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

Answer: (B)

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