Question

Which of the following is the possible energy of an excited state of hydrogen atom?

• A. $-3.4\text{eV}$
• B. $+6.8\text{eV}$
• C. $-13.6\text{eV}$
• D. $-6.8\text{eV}$

Answer 1

The correct option is A $-3.4\text{eV}$

Total energy of an electron in an orbit is given by,

$E=\left(\frac{Z^2}{n^2}\right)(-13.6\text{eV})$

Where, $Z$ is an atomic number. For hydrogen atom $Z=1.$

$n=1,2,3,\mathrm{4.......}$

$n=1\to \text{ground state}$
$n=2\to \text{first excited state}$
$n=3\to \text{second excited state}$

For $n=1$ we get,
$E_1=-13.6\text{eV}$

For $n=2$ we get,
$E_2=\frac{-13.6}{2^2}=-3.4\text{eV}$

Here, only the energy of the excited state is asked. So, option $\left(C\right)$ cannot be the answer,

Hence, option $\left(A\right)$ is correct.

Why this question? Key Note: The negative sign indicates that a negative work has to be done to remove the electron from the bound of the atom to infinity, where it is considered to have zero energy.