Question

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

• 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 $E=3.4eV,\lambda =6.6\times {10}^{-10}m$

E=-(Total energy)
=3.4 eV
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$
$=\frac{(6.6\times {10}^{-34})}{\sqrt{2\times 1.6\times {10}^{-19}\times \left(3.4eV\right)}}$
$=6.6\times {10}^{-10}m$