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 wavelength of the emitted radiation when the electron drops from third to second orbit is:

• A. $4.6\times {10}^6$
• B. $6.6\times {10}^3$
• C. $0.66\times {10}^6$
• D. none of these

Answer 1

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

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