Question

According to Bohr's theory, which of the following transactions in the hydrogen atom will rise to the least energetic photon?
A) $\text{ n = 6 }$ To $\text{ n = 1 }$
B) $\text{ n = 5 }$ To $\text{ n = 4 }$
C) $\text{ n = 6 }$ To $\text{ n = 5 }$
D) $\text{ n = 5 }$ To $\text{ n = 3 }$

Answer 1

Hint: Let electron jumps from energy level $\text{ }{{\text{n}}_{\text{1}}}\text{ }$ with energy $\text{ }{{\text{E}}_{\text{1}}}\text{ }$ to the energy level $\text{ }{{\text{n}}_{2}}\text{ }$ where energy $\text{ }{{\text{E}}_{2}}\text{ }$ . Then the energy difference between the two energy level is given by,
$\text{ }\Delta \text{E = }{{\text{E}}_{\text{2}}}\text{ }-{{\text{E}}_{\text{1}}}\text{ = }{{\text{R}}_{\text{H}}}{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\text{ }$
Where, $\text{ }{{\text{R}}_{\text{H}}}\text{ }$ is a Rydberg constant, z is the atomic number of atoms.

Complete step by step answer:
According to the Bohr Theory, on the exposure of the energy, the solitary electrons in the different energy levels will absorb the energy and shift to the different (higher) energy level depending on the energy absorbed by the atoms. Some may absorb the energy to shift their electron to the second level some may shift third or even higher level.
The electron in the excited state de-excite to the lower energy levels and releases the energy in terms of radiation of specific frequency.
Let electron jumps from energy level $\text{ }{{\text{n}}_{\text{1}}}\text{ }$ with energy $\text{ }{{\text{E}}_{\text{1}}}\text{ }$ to the energy level $\text{ }{{\text{n}}_{2}}\text{ }$where energy $\text{ }{{\text{E}}_{2}}\text{ }$ . Then the energy difference between the two energy level is given by,
$\text{ }\Delta \text{E = }{{\text{E}}_{\text{2}}}\text{ }-{{\text{E}}_{\text{1}}}\text{ = }{{\text{R}}_{\text{H}}}{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\text{ }$
Where, ${{\text{R}}_{\text{H}}}$ is a Rydberg constant, z is the atomic number of atoms.
According to which, the difference in energies is inversely related to the energy level n1 and n2.The least energetic photon would be released when the difference in the energies level is least.
Let's have a look at the option.
a) Energy for the excitation of an electron from $\text{ n = 6 }$ to $\text{ n = 1 }$
Here, $\text{ }{{\text{n}}_{\text{1}}}\text{ = 1 }$and $\text{ }{{\text{n}}_{2}}\text{ = 6 }$
Then the energy of the photon would be,
$\begin{align}
 & \text{ }\Delta \text{E = }{{\text{R}}_{\text{H}}}{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{1}-\dfrac{1}{{{6}^{2}}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{1}-\dfrac{1}{36} \right) \\
 & \therefore \Delta \text{E= }{{\text{R}}_{\text{H}}}\left( \dfrac{35}{36} \right)=0.97{{\text{R}}_{\text{H}}} \\
\end{align}$
b) Energy for the excitation of an electron from $\text{ n = 5 }$ to $\text{ n = 4 }$
Here, $\text{ }{{\text{n}}_{\text{1}}}\text{ = 4 }$and $\text{ }{{\text{n}}_{2}}\text{ = 5 }$
Then the energy of the photon would be,
$\begin{align}
  & \text{ }\Delta \text{E = }{{\text{R}}_{\text{H}}}{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{5}^{2}}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{16}-\dfrac{1}{25} \right) \\
 & \therefore \Delta \text{E= }{{\text{R}}_{\text{H}}}\left( \dfrac{9}{400} \right)=0.022{{\text{R}}_{\text{H}}} \\
\end{align}$
c) Energy for the excitation of an electron from $\text{ n = 6 }$ to the $\text{ n = 5 }$
Here, $\text{ }{{\text{n}}_{\text{1}}}\text{ = 5 }$and $\text{ }{{\text{n}}_{2}}\text{ = 6 }$
Then the energy of the photon would be,
$\begin{align}
  & \text{ }\Delta \text{E = }{{\text{R}}_{\text{H}}}{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{{{5}^{2}}}-\dfrac{1}{{{6}^{2}}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{25}-\dfrac{1}{36} \right) \\
 & \therefore \Delta \text{E= }{{\text{R}}_{\text{H}}}\left( \dfrac{11}{900} \right)=\text{ 0}\text{.01 }{{\text{R}}_{\text{H}}} \\
\end{align}$
d) Energy for the excitation of an electron from $\text{ n = 5 }$ to $\text{ n = 3 }$
Here, $\text{ }{{\text{n}}_{\text{1}}}\text{ = 3 }$and $\text{ }{{\text{n}}_{2}}\text{ = 5 }$

Then the energy of the photon would be,
$\begin{align}
  & \text{ }\Delta \text{E = }{{\text{R}}_{\text{H}}}{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{5}^{2}}} \right)\text{ } \\
 & \Rightarrow \text{= }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{9}-\dfrac{1}{25} \right) \\
 & \therefore \Delta \text{E= }{{\text{R}}_{\text{H}}}\left( \dfrac{16}{225} \right)=\text{ 0}\text{.07}{{\text{R}}_{\text{H}}} \\
\end{align}$

Here, the excitation of an electron from the level $\text{ }{{\text{n}}_{2}}\text{ = 6 }$ to the $\text{ }{{\text{n}}_{\text{1}}}\text{ = 5 }$ release of the less energetic photon. This difference is less as compared to the other transitions.

So, the correct answer is “Option C”.

Note: Note that, here we are dealing with the hydrogen atom thus the energy of the hydrogen atom is also written as $\text{ E = }-13.6\dfrac{{{Z}^{2}}}{{{\text{n}}^{\text{2}}}}eV\text{ }$ where the energy is inversely related to the number of energy level.$\text{ E }\propto \text{ }\dfrac{1}{{{\text{n}}^{\text{2}}}}\text{ }$. The wave number $\text{ }\overset{-}{\mathop{\nu }}\,\text{ }$ of the spectral line or the emitted radiation can be given by the formula:
$\text{ }\overset{-}{\mathop{\nu }}\,\text{ = RH}\left( \dfrac{1}{\text{n}_{1}^{2}}-\dfrac{1}{\text{n}_{2}^{2}} \right)\text{ }$
Less is the wavenumber higher is the less is the frequency of the radiation and thus, the less energetic photon.