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?

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Answer 1 · 2015-12-08T08:22:31

$=>lambda=102.6nm$

Energy is absorbed.

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

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

$DeltaE=-2.178xx10^(-18)Z^(2)J(1/(n_("final")^2)-1/(n_("initial")^2))$

Here, the initial level is $n_("initial")=1$ and the final level is $n_("final")=3$ . Since this is hydrogen, then $Z=1$ .

Thus, $DeltaE=-2.178xx10^(-18)(1)^(2)J(1/(9)-1/(1))=1.936xx10^(-18)J$

This energy can also be given as $DeltaE=hnu=hc/(lambda)$

$=>lambda=hc/(DeltaE)=(6.626xx10^(-34)xx2.998xx10^8)/(1.936xx10^(-18))=1.026xx10^(-7)m$

$=>lambda=102.6nm$

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 $DeltaE$ is positive ( $DeltaE$ >0).

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