Question

In a Bohr's model of an atom, when an electron jumps from $n=1$ to $n=3$, how much energy will be emitted or absorbed?

• A. $2.389\times {10}^{-12}$ j
• B. $0.239\times {10}^{-10}$ j
• C. $1.94\times {10}^{-10}$ j
• D. $0.1936\times {10}^{-10}$ j

Answer 1

The correct option is A $2.389\times {10}^{-12}$ j

$\Delta E=(-R_H/n_f^2)-(-R_H/n_i^2)$

$n_i$ & $n_f$ stands for initial and final orbit respectively

$R_H$=rydrberg's constant $=2.18\times {10}^{-18}$

$\Delta E=(-2.18\times {10}^{-18}/3^2)-(-2.18\times {10}^{-18}/1^2)$

$\Delta E=2.18\times {10}^{-18}(1/3^2-1)$

$\Delta E=1.94\times {10}^{-10}J$