Question

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

• A. $6.6\times {10}^{-5}$
• B. $6.6\times {10}^{-7}$
• C. $3\times {10}^{-7}$
• D. $3\times {10}^{-5}$

Answer 1

The correct option is A $6.6\times {10}^{-5}$

We have,
$\Delta E=E_3-E_2$
$\Delta E=-2.41\times {10}^{-12}-(-5.42\times {10}^{-12})erg$
$\Delta E=3.01\times {10}^{-12}erg$
$\Delta E=\frac{hc}{\lambda }$
$\lambda =\frac{hc}{\Delta E}$
$\lambda =\frac{6.6\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{3.01\times {10}^{-12}erg}$
$\lambda =\frac{6.6\times {10}^{-34}\times {10}^{7}ergs\times 3\times {10}^{10}cm{s}^{-1}}{3.01\times {10}^{-12}erg}$
$\lambda =6.6\times {10}^{-5}cm$