Question

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

• A. $6.604\times {10}^3\stackrel{\circ }{A}$
• B. $6.604\stackrel{\circ }{A}$
• C. $3.905\times {10}^3\stackrel{\circ }{A}$
• D. $3.905\stackrel{\circ }{A}$

Answer 1

The correct option is A $6.604\times {10}^3\stackrel{\circ }{A}$

Energy of an electron in the n${}^{th}$ orbit is denoted as $E_n$. So, the energies of orbits 3 and 2 can be written as $E_3$ and $E_2$ respectively.
$\Delta E$ is the difference between these energies.
If $\lambda$ is the wavelength of the emitted radiation, we can say that:
$\Delta E=E_3-E_2=\frac{hc}{\lambda }$
$⟹\lambda =\frac{hc}{E_3-E_2}$
Given $E_2=-5.42\times {10}^{-12}erg$ and $E_3=-2.41\times {10}^{-12}erg$
We know that $c=3\times {10}^{10}cm/s.$
$\therefore \lambda =\frac{6.626\times {10}^{-27}\times 3\times {10}^{10}}{-2.41\times {10}^{-12}-(-5.42\times {10}^{-12})}$= $\frac{19.878\times {10}^{-17}}{3.01\times {10}^{-12}}=6.604\times {10}^{-5}cm=6.604\times {10}^3\stackrel{\circ }{A}$