Question

An electron in hydrogen like atom is in excited state. It has a total energy of $–3.4\text{eV}$. For the electron,

• A. Kinetic energy is $6.8\text{eV}$
• B. de Broglie wavelength is 6.63 $\underset{A}{\circ }$
• C. Potential energy is $-6.8\text{eV}$
• D. Kinetic energy is $3.4\text{eV}$

Answer 1

Answer: (B), (C), (D)

Kinetic energy is $3.4\text{eV}$

$\text{Kinetic energy}=\text{Total energy}=3.4\text{eV}$
$=-\frac{\text{Potential energy}}{2}$
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m\times K.E.}}=6.63\underset{A}{\circ }$