Question

The energy of an electron in the second and the third Bohr's orbits of a hydrogen atom is $-5.42\times {10}^{-12}\text{erg}$ and $-2.41\times {10}^{-12}\text{erg}$ respectively. Calculate the wavelength of the emitted radiation when the electron drops from the third to the second orbit.

• A. $5603\stackrel{o}{\text{A}}$
• B. $6603\stackrel{o}{\text{A}}$
• C. $7603\stackrel{o}{\text{A}}$
• D. $8603\stackrel{o}{\text{A}}$

Answer 1

The correct option is B $6603\stackrel{o}{\text{A}}$

For transition emitted from $n=3$ to $n=2$, energy is give by,
$\Delta E=E_3-E_2=(-2.41\times {10}^{-12}-(-5.42\times {10}^{-12}\left)\right)\text{erg}$
$\Rightarrow \frac{hc}{\lambda }$

$\Rightarrow \lambda =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{3.01\times {10}^{-19}}$

$=6603\stackrel{o}{\text{A}}$