
How much energy is required to excite a hydrogen atom from its ground state to the second excited state?
Answer
133.8k+ views
Hint: We know that there are different energy states for atoms. Atoms are most likely to remain in their ground state. But if we provide enough energy for the atoms will jump from its ground state to the excited states. Here we have to find the energy required to excite the hydrogen atom from its ground state to its second excited state. Atoms are in an excited state when the electrons absorb energy and jump to higher energy levels.
Formula used
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
Where ${E_n}$ is the energy of the hydrogen atom in the ${n^{th}}$ energy level and $n$ stands for the required energy level to which we excite the atom. We get the constant value of $13.6$ through calculation.
Complete step by step solution:
We know that the energy of the hydrogen atom at the ${n^{th}}$ energy level is given by,
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
The energy required to excite the hydrogen atom from the ground state to its second excited state will be equal to the difference between the energies of the excited state and the ground state of the hydrogen atom,
i.e.
For ground state,$n = 1$
For the second excited state,$n = 3$
The difference in energy can be written as,
${E_n} = \dfrac{{ - 13.6}}{{{3^2}}} - \dfrac{{ - 13.6}}{{{1^2}}}$
Taking common values outside, we get
${E_n} = - 13.6\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{1^2}}}} \right)$
This can be written as,
${E_n} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right)$
$ \Rightarrow {E_n} = 12.08eV$
The answer is: $12.08eV$
Note:
In the expression for the energy of the ${n^{th}}$energy level, there is a negative sign. This negative sign shows that the electron is bound to the nucleus. If we can provide an energy of $\dfrac{{13.6}}{{{n^2}}}eV$ to the electron located in the ${n^{th}}$orbit, it can escape to a point infinitely far away from the nucleus of the atom. The lowest energy level of an atom is called its ground state.
Formula used
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
Where ${E_n}$ is the energy of the hydrogen atom in the ${n^{th}}$ energy level and $n$ stands for the required energy level to which we excite the atom. We get the constant value of $13.6$ through calculation.
Complete step by step solution:
We know that the energy of the hydrogen atom at the ${n^{th}}$ energy level is given by,
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
The energy required to excite the hydrogen atom from the ground state to its second excited state will be equal to the difference between the energies of the excited state and the ground state of the hydrogen atom,
i.e.
For ground state,$n = 1$
For the second excited state,$n = 3$
The difference in energy can be written as,
${E_n} = \dfrac{{ - 13.6}}{{{3^2}}} - \dfrac{{ - 13.6}}{{{1^2}}}$
Taking common values outside, we get
${E_n} = - 13.6\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{1^2}}}} \right)$
This can be written as,
${E_n} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right)$
$ \Rightarrow {E_n} = 12.08eV$
The answer is: $12.08eV$
Note:
In the expression for the energy of the ${n^{th}}$energy level, there is a negative sign. This negative sign shows that the electron is bound to the nucleus. If we can provide an energy of $\dfrac{{13.6}}{{{n^2}}}eV$ to the electron located in the ${n^{th}}$orbit, it can escape to a point infinitely far away from the nucleus of the atom. The lowest energy level of an atom is called its ground state.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

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

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage

Current Loop as Magnetic Dipole and Its Derivation for JEE
