Atomic Structure - Result Question 50
####26. The energy of an electron in the first Bohr orbit of H-atom is $-13.6 eV$. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are) $\quad(1998,2 M)$
(a) $-3.4 eV$
(b) $-4.2 eV$
(c) $-6.8 eV$
(d) $+6.8 eV$
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Solution:
- The energy of an electron in a Bohr atom is expressed as
$$ \begin{aligned} E _n=-\frac{k Z^{2}}{n^{2}} \quad \text { where, } \begin{aligned} k & =\text { Constant } \ Z & =\text { Atomic number, } \ n & =\text { Orbit number } \ & =-13.6 eV \text { for } H(n=1) \end{aligned} \end{aligned} $$
when $n=2, E _2=\frac{-13.6}{2^{2}} eV=-3.40 eV$
( $n$ can have only integral value $1,2,3, \ldots \ldots \infty$ )
