Question

The energy of the electron in the second and third Bohr orbits of the hydrogen atom is -5.42 $\times$ 10${}^{-12}$ erg and -2.41 $\times$ 10${}^{-12}$ erg, respectively. The wave length of the emitted radiation when the electron drops from third to second orbit will be :

• A. 6.6 $\times$ 10${}^3$
• B. 6.6 $\times$ 10${}^{-3}$
• C. 13.2 $\times$ 10${}^3$
• D. 13.2 $\times$ 10${}^{-3}$

Answer 1

The correct option is A 6.6 $\times$ 10${}^3$

$\Delta$E = -2.41 $\times$ 10$6-12$ -(-5.42 $\times$ 10$6-12$)
= 3.01 $\times$ 10${}^{-12}$ erg
$\Delta$E = hv $\Rightarrow$ $\lambda$= $\frac{c}{v}$ = $\frac{3\times {10}^{10}}{3.01\times {10}^{-12}/(6.62\times 106-27}$
= 6.6 $\times$ 10${}^{-5}$
= 6.6 $\times$ 10${}^{-5}$ $\times$ 10${}^8$
=$6.6\times$ 10${}^3$