
The energy of the electron in the first Bohr’s orbit of the H atom is -13.6 eV. What will be its potential energy in n = 4th orbit?
(a) -13.6 eV
(b) -3.4 eV
(c) -0.85 eV
(d) -1.70 eV
Answer
160.8k+ views
Hint: We know that Bohr's theory was proposed for the hydrogen atom which says that the movement of electrons around the nucleus takes place only in specific orbits. And when the electrons jump to an orbit of lower energy, emission of radiation takes place.
Formula used: The formula for the energy possessed by an electron in the nth orbit is,
\[{E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV\]. Here, n represents the orbit number.
Complete Step by Step Answer:
Here, first, we will find the electron's energy in the 4th orbit of the atom of hydrogen.
\[{E_n} = \dfrac{{ - 13.6}}{{{4^2}}}eV = - 0.85\,eV\]
So, the total energy in the 4th orbit is -0.85 eV.
Now, we have to find the relation between total energy and potential energy. In the atom of hydrogen, potential energy is two times the total energy in the nth orbit of the electron.
So, the relation is,
\[{E_P} = 2{E_n}\]
Here, \[{E_P}\] stands for potential energy, and \[{E_n}\] stands for energy at the nth orbit.
We got, \[{E_n} = - 0.85\,eV\]
Therefore,
\[{E_P} = 2{E_n}\]
\[{E_P} = 2 \times - 0.85 = - 1.70\,{\rm{eV}}\]
Therefore, the electron's potential energy is -1.70 eV.
Hence, the correct option is (D).
Note: It is to be noted that, the total energy of an electron in higher orbits decreases with the increase of the number of orbits. It is because of the fact that the electrons of the higher energy orbits are not strongly attracted by the nucleus. Thus, the removal of the electrons is easy.
Formula used: The formula for the energy possessed by an electron in the nth orbit is,
\[{E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV\]. Here, n represents the orbit number.
Complete Step by Step Answer:
Here, first, we will find the electron's energy in the 4th orbit of the atom of hydrogen.
\[{E_n} = \dfrac{{ - 13.6}}{{{4^2}}}eV = - 0.85\,eV\]
So, the total energy in the 4th orbit is -0.85 eV.
Now, we have to find the relation between total energy and potential energy. In the atom of hydrogen, potential energy is two times the total energy in the nth orbit of the electron.
So, the relation is,
\[{E_P} = 2{E_n}\]
Here, \[{E_P}\] stands for potential energy, and \[{E_n}\] stands for energy at the nth orbit.
We got, \[{E_n} = - 0.85\,eV\]
Therefore,
\[{E_P} = 2{E_n}\]
\[{E_P} = 2 \times - 0.85 = - 1.70\,{\rm{eV}}\]
Therefore, the electron's potential energy is -1.70 eV.
Hence, the correct option is (D).
Note: It is to be noted that, the total energy of an electron in higher orbits decreases with the increase of the number of orbits. It is because of the fact that the electrons of the higher energy orbits are not strongly attracted by the nucleus. Thus, the removal of the electrons is easy.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Degree of Dissociation and Its Formula With Solved Example for JEE
