Question

The energy of an electron in the first Bohr orbit of $H$-atom is $–13.6eV$. The possible energy value(s) of the excited state(s) for electrons in Bohr's orbits of hydrogen is (are):

• A. $–3.4eV$
• B. $–4.2eV$
• C. $–6.8eV$
• D. $+6.8eV$

Answer 1

The correct option is A $–3.4eV$

The energy of an electron in a Bohr atom is expressed as
$E=\frac{-k{Z}^2}{n^2}$
where, $k$ = Constant,
$Z$ = Atomic number,
$n$ = Orbit number
= $-13.6eV$ for $H(n=1)$
when $n=2$ , $E_2=\frac{-13.6}{22}eV=-3.40eV$
(Where $n$ can have only integral value $1,2,3,\dots \dots \infty )$
For higher integral values of $n(n>2)$, the value of $E$ will be lesser in magnitude as compared to $3.40eV$. So none of the options will match.