Q. The ionization energy of hydrogen atom is $13.6\ eV$. Following Bohr’s theory, the energy corresponding to a transition between the 3rd and the 4th orbit is
A) $3.40\ eV$
B) $1.51\ eV$
C) $0.85\ eV$
D) $0.66\ eV$
Explanation
(d) $E_{3} = - \frac{13.6}{9} = - 1.51eV;E_{4} = - \frac{13.6}{16} = - 0.85eV$
$\therefore E_{4} - E_{3} = 0.66\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.