A photon on a hydrogen atom excited an electron from n = 1 to make a transition to n=3. What is the wavelength (in run) pertaining to this transition and was the transition emitting or absorbing in nature?

Dec 8, 2015

$\implies \lambda = 102.6 n m$

Energy is absorbed.

Explanation:

In order to solve this question we will use the Bohr Model.

The change in energy according to Bohr can be calculated by:

$\Delta E = - 2.178 \times {10}^{- 18} {Z}^{2} J \left(\frac{1}{{n}_{\text{final")^2)-1/(n_("initial}}^{2}}\right)$

Here, the initial level is ${n}_{\text{initial}} = 1$ and the final level is ${n}_{\text{final}} = 3$. Since this is hydrogen, then $Z = 1$.

Thus, $\Delta E = - 2.178 \times {10}^{- 18} {\left(1\right)}^{2} J \left(\frac{1}{9} - \frac{1}{1}\right) = 1.936 \times {10}^{- 18} J$

This energy can also be given as $\Delta E = h \nu = h \frac{c}{\lambda}$

$\implies \lambda = h \frac{c}{\Delta E} = \frac{6.626 \times {10}^{- 34} \times 2.998 \times {10}^{8}}{1.936 \times {10}^{- 18}} = 1.026 \times {10}^{- 7} m$

$\implies \lambda = 102.6 n m$

Since the electron is moving from a lower energy level to a higher energy level, the energy is absorbed.

Moreover, we can say that the energy is absorbed because the sign of $\Delta E$ is positive ($\Delta E$>0).