
Using Bohr's equation for the energy levels of the electron in a Hydrogen atom, determine the energy of an electron in \[n\text{ }=\text{ }4\]
A) $-5.45\times {{10}^{-19}}J$
B) $-1.84\times {{10}^{-29}}J$
C) $-1.365\times {{10}^{-19}}J$
D) $1.84\times {{10}^{-29}}J$
Answer
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Hint: In the Bohr model of an atom, the electrons travel in the circular orbits around the nucleus that are defined. These orbits are labeled as the quantum number $n$ The electrons jump from one orbit to another by emitting energy. The formula for calculating the energy of electron is-
\[{{E}_{n}}\text{ }=\text{ }-\text{ }\left( \text{ }\dfrac{e{}^\text{2}}{8\cdot \pi \cdot {{e}_{0}}\cdot n{}^\text{2}\cdot {{a}_{0}}} \right)\text{ }and\text{ }{{E}_{n}}\text{ }=\text{ }-\left( \dfrac{13.6}{n{}^\text{2}} \right)~ev\]
Complete step-by-step answer:
Energy of an electron in the first Bohr orbit of \[H\text{ }atom\text{ }=\text{ }-13.6eV\]
The energy value of the excited state for electron and the formula to calculate the energy of an electron is
\[{{E}_{n}}\text{ }=\text{ }-\text{ }\left( \text{ }\dfrac{e{}^\text{2}}{8\cdot \pi \cdot {{e}_{0}}\cdot n{}^\text{2}\cdot {{a}_{0}}} \right)\]
The energy obtained is always a negative number and the ground state \[n\text{ }=\text{ }1\] has the most negative value. The reason being that the energy of an electron in orbit is relative to the energy of an electron that is entirely separated from its nucleus \[~n=\infty ~\] and it is recognised to have an energy of \[0\text{ }eV\] Since the electron in a fixed orbit around the nucleus is more stable than an electron that is extremely far from its nucleus, the energy of the electron in orbit is always negative.
\[\Rightarrow \text{ }{{E}_{n}}\text{ }=\text{ }-\left( \dfrac{{{E}_{1}}}{n{}^\text{2}} \right)~ev\]
\[\Rightarrow \text{ }{{E}_{n}}\text{ }=\text{ }-\left( \dfrac{13.6}{n{}^\text{2}} \right)~ev\]
${\Rightarrow {E}_{4}}=-\left( \dfrac{13.6}{{{n}^{2}}} \right)\times 1.6\times {{10}^{-19}}$
${\Rightarrow {E}_{4}}=-1.365\times {{10}^{-19}}$
Therefore, Option C is correct answer i.e. the energy of an electron in \[n\text{ }=\text{ }4\] is $-1.365\times {{10}^{-19}}$
Note: A student can get confused between ground state and excited state of an electron. Ground state: The ground state of an electron is the energy level of the electron that it usually occupies. The ground state is the lowest energy state of the electron.
Excited state: The excited state of an electron is the energy state which the electron temporarily acquires. This state is greater than the ground state.
\[{{E}_{n}}\text{ }=\text{ }-\text{ }\left( \text{ }\dfrac{e{}^\text{2}}{8\cdot \pi \cdot {{e}_{0}}\cdot n{}^\text{2}\cdot {{a}_{0}}} \right)\text{ }and\text{ }{{E}_{n}}\text{ }=\text{ }-\left( \dfrac{13.6}{n{}^\text{2}} \right)~ev\]
Complete step-by-step answer:
Energy of an electron in the first Bohr orbit of \[H\text{ }atom\text{ }=\text{ }-13.6eV\]
The energy value of the excited state for electron and the formula to calculate the energy of an electron is
\[{{E}_{n}}\text{ }=\text{ }-\text{ }\left( \text{ }\dfrac{e{}^\text{2}}{8\cdot \pi \cdot {{e}_{0}}\cdot n{}^\text{2}\cdot {{a}_{0}}} \right)\]
The energy obtained is always a negative number and the ground state \[n\text{ }=\text{ }1\] has the most negative value. The reason being that the energy of an electron in orbit is relative to the energy of an electron that is entirely separated from its nucleus \[~n=\infty ~\] and it is recognised to have an energy of \[0\text{ }eV\] Since the electron in a fixed orbit around the nucleus is more stable than an electron that is extremely far from its nucleus, the energy of the electron in orbit is always negative.
\[\Rightarrow \text{ }{{E}_{n}}\text{ }=\text{ }-\left( \dfrac{{{E}_{1}}}{n{}^\text{2}} \right)~ev\]
\[\Rightarrow \text{ }{{E}_{n}}\text{ }=\text{ }-\left( \dfrac{13.6}{n{}^\text{2}} \right)~ev\]
${\Rightarrow {E}_{4}}=-\left( \dfrac{13.6}{{{n}^{2}}} \right)\times 1.6\times {{10}^{-19}}$
${\Rightarrow {E}_{4}}=-1.365\times {{10}^{-19}}$
Therefore, Option C is correct answer i.e. the energy of an electron in \[n\text{ }=\text{ }4\] is $-1.365\times {{10}^{-19}}$
Note: A student can get confused between ground state and excited state of an electron. Ground state: The ground state of an electron is the energy level of the electron that it usually occupies. The ground state is the lowest energy state of the electron.
Excited state: The excited state of an electron is the energy state which the electron temporarily acquires. This state is greater than the ground state.
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