
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
544.2k+ views
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is Environment class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

How many squares are there in a chess board A 1296 class 11 maths CBSE

Distinguish between verbal and nonverbal communica class 11 english CBSE

The equivalent weight of Mohrs salt FeSO4 NH42SO4 6H2O class 11 chemistry CBSE

