   # An excited hydrogen atom emits light with a wavelength of 397.2 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 2, Problem 68E
Textbook Problem
931 views

## An excited hydrogen atom emits light with a wavelength of 397.2 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin?

Interpretation Introduction

Interpretation:

The wavelength and final state of an electronic transition in Hydrogen atom is given. The initial excite state of the electron is to be determined.

Concept introduction:

According to the quantum mechanics, an electron in Hydrogen atom can occupy certain states corresponds to certain energy levels. The energy of these levels inversely depends on the square of principle quantum number n .

Rydberg propose a formula to calculate the energy or wavelength of the emitted photon when a transition takes place in Hydrogen atom.

To determine: Initial excited state for the electron.

### Explanation of Solution

To determine: The energy change during the transition.

Given

Final excited state, nfinal=2

The conversion of manometer (nm) into meter (m) is done as,

1nm=109m

Hence, the conversion of 397.2nm into centimeter is,

397.2nm=397.2×109m=3.972×107m

Calculate the energy change during the ionization of atom.

ΔE=hcλ

Where,

• λ is the wavelength.
• h is the Plank’s constant (6.62×1034Js) .
• c is the speed of light (3×108m/s) .
• ΔE is the energy change for the ionization.

Substitute the values of h,c and λ in the above equation.

ΔE=hcλ=(6.62×1034Js)(3×108m/s)3

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