Question

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

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

In any state,
$KE=E,andPE=-2E$, hence the total energy is $-E.$
Here $E=3.4eV.$
Converting $E$ into $Joules$ and using the relation.

$E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{E}=6.6\ast {10}^{-10}m$