Question

The ionization energy of the hydrogen atom is $13.6eV$. Following Bohr's theory, the energy corresponding to a transition between the $3^{rd}$ and $4^{th}$ orbit is:

• A. $3.40eV$
• B. $1.51eV$
• C. $0.85eV$
• D. $0.66eV$

Answer 1

The correct option is D $0.66eV$

According to Bohr's theory,

$E_n=\frac{-13.6}{n^2}$

$\therefore E_3=\frac{-13.6}{3^2}=\frac{-13.6}{9}=-1.51eV$

$E_4=\frac{-13.6}{4^2}=\frac{-13.6}{16}=-0.85eV$

Energy corresponding to a transition between the $3^{rd}$ and $4^{th}$ orbit $=E_4-E_3=-0.85-(-1.51)=0.66eV$

Hence, the energy corresponding to a transition between the $3^{rd}$ and $4^{th}$ orbit is $0.66eV$.