Q.58 Energy of the electron in hydrogen atom in its ground state is 13.6 eV. The energy required (in eV) to move the electron from its ground state to the first excited state, rounded off to TWO decimal places, is ________.
Q.58

Energy of the electron in hydrogen atom in its ground state is 13.6 eV. The energy required (in eV) to move the electron from its ground state to the first excited state, rounded off to TWO decimal places, is ________.

Energy Required: 10.20 eV

The energy required to excite the electron in a hydrogen atom from the ground state (n=1) to the first excited state (n=2) is 10.20 eV.

Bohr Model Basics

The Bohr model describes electron energy levels in hydrogen as \( E_n = -\frac{13.6}{n^2} \) eV, where n is the principal quantum number. Ground state energy at n=1 is \( E_1 = -13.6 \) eV.First excited state at n=2 gives \( E_2 = -\frac{13.6}{4} = -3.4 \) eV.

Energy Calculation

Energy difference is \( \Delta E = E_2 – E_1 = -3.4 – (-13.6) = 10.2 \) eV.[web:5][web:6][web:12][code:0] Rounded to two decimal places, this is 10.20 eV. Positive value indicates energy absorbed for excitation.

Common Errors

Some confuse this with ionization energy (13.6 eV to remove electron entirely).[web:6] Others miscalculate \( E_2 \) as -13.6/2 instead of -13.6/4.[web:12] Transition from n=1 to n=2 specifically yields 10.2 eV.


Hydrogen Atom Ground State to First Excited State Energy: Complete Guide

The hydrogen atom ground state to first excited state energy transition is a cornerstone of atomic physics, vital for CSIR NET Life Sciences and competitive exams. With ground state energy at -13.6 eV, the energy required to move the electron to the first excited state reveals quantized levels from Bohr’s model.

Bohr Energy Formula Explained

Electrons occupy discrete orbits where \( E_n = -\frac{13.6}{n^2} \) eV.[web:4][web:14] For n=1 (ground state), \( E_1 = -13.6 \) eV; for n=2 (first excited), \( E_2 = -3.4 \) eV.[web:5][code:0] Excitation energy is the absolute difference: 10.20 eV.

Step-by-Step Solution

  1. Compute \( E_1 = -\frac{13.6}{1^2} = -13.60 \) eV.
  2. Compute \( E_2 = -\frac{13.6}{2^2} = -3.40 \) eV.
  3. \( \Delta E = |E_2 – E_1| = 10.20 \) eV (rounded).

This matches exam expectations, avoiding pitfalls like using n=3 levels.

Exam Relevance for CSIR NET

CSIR NET questions test this exact hydrogen atom ground state to first excited state energy (10.20 eV fill-in).[web:6] Understand Balmer series links (n=2 base) for spectra. Practice confirms no options needed—direct numerical answer.

Quick Reference Table

State n Energy (eV)
Ground 1 -13.60
First Excited 2 -3.40
ΔE Required 1→2 10.20 eV

 

Leave a Reply

Your email address will not be published. Required fields are marked *

Latest Courses