Question

An electron changes from \[n=2\] to \[n=6\] energy state. What is the energy of the photon in joules?

Answer 1

Hint: We know that we use the formula for energy of the photon in terms of wavelength of the photon. Using this formula calculate the energy of the photon emitted. Also determine the energy levels in which the transition of the electron takes place. Use the formula for change in angular momentum of the atom and calculate the required change in momentum of the hydrogen atom.

Complete answer:
In more technical terms we will describe ionization energy the minimum energy that an electron during a gaseous atom or ion needs to absorb to return out of the influence of the nucleus. Normally, when the ionization energy is high it'll be harder to get rid of an electron. There also are several factors that govern the attraction forces.
If the nucleus is charged then the electrons are strongly interested in it.
If an electron lies near or on the brink of the nucleus then the attractions are going to be greater than the one when the electron is further away.
If there are more electrons between the outer level and therefore the nucleus the attraction forces are less.
When there are two electrons within the same orbital they experience some sort of repulsion. Now, this creates disturbances within the attraction of the nucleus. In essence, ionization energy are going to be less in paired electrons as they will be removed easily.
Here we have \[{{n}_{i}}=2\]to \[{{n}_{f}}=6\] [R is the Rydberg constant, equal to $1.097\times {{10}^{7}}{{m}^{-1}}$ ]
When the electron falls from \[{{n}_{i}}=6\]to \[{{n}_{f}}=2\] by using the Rydberg equation.
\[\dfrac{1}{\lambda }=R\times \left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)\]
Substituting the values in above equation;
\[\dfrac{1}{\lambda }=1.097\times {{10}^{7}}\times \left( \dfrac{1}{2_{{}}^{2}}-\dfrac{1}{6_{{}}^{2}} \right)\]
On further solving we get;
\[\dfrac{1}{\lambda }=2.4387\times {{10}^{6}}{{m}^{-1}}\Rightarrow \lambda =4.10\times {{10}^{-7}}m\]
So, now we know that when an electron falls from \[{{n}_{i}}=2\]to \[{{n}_{f}}=6\] it must absorb a photon of the same wavelength.
$E=h\times \dfrac{c}{\lambda }$ here we know that $c=3\times {{10}^{8}}m{{s}^{-1}}$ and $h=6.626\times {{10}^{-34}}Js$
Substituting the values in above equation;
$E=6.626\times {{10}^{-34}}\times \dfrac{3\times {{10}^{8}}}{4.10\times {{10}^{-7}}}$
On further solving we get;
\[E=4.85\times {{10}^{-19}}J\]
Therefore, an electron changes from an \[n=2\] to an \[n=6\] energy state. The energy of the photon in joules is \[4.85\times {{10}^{-19}}J\]

Note:
You may think that how can we say that the electron jumps from the third energy level to the first energy level and not from the first to third energy level. The first thing is that the photon is emitted which shows that the atom has excess energy and also the energy of the photon is positive. If the electron was jumping from first to third energy level then it has to absorb the energy and this energy would have been negative.