Question

An electric in the first orbit of hydrogen atom, has a binding energy of $13.6\text{eV}$. If an electron in another hydrogen atom is in the $n=3$ state, how much energy is required to ionize it?

• A. $13.6\text{eV}$
• B. $4.53\text{eV}$
• C. $3.4\text{eV}$
• D. $1.51\text{eV}$

Answer 1

The correct option is D $1.51\text{eV}$

The binding energy of an electron is,
$B.E.=\frac{13.6}{n^2}\left(\text{in eV}\right)$
The energy required to ionize the hydrogen atom is equal to the binding energy of an electron in the third orbit.
So, $E_3=\frac{+13.6}{n^2}$

$\Rightarrow E_3=\frac{13.6}{3^2}=1.51\text{eV}$

Hence, $\left(D\right)$ is the correct answer.