Question

Of the following transitions in the Bohr hydrogen atom, the __________ transition results in the emission of the highest-energy photon.
A. n=6 to n=1
B. n=1 to n=6
C. n=1 to n=5
D. n=5 to n=1

Answer 1

Answer:

• Option A. n = 6 to n = 1

Explanation:

Niels Bohr was a Danish physicist who proposed the hydrogen atom quantum model to explain the discontinuity of the atom's emission spectra.

In Bohr hydrogen atom model, the electrons occupy orbits identified with the numbers n = 1, 2, 3, 4, ... Each number (orbit) corresponds to a different energy level  or state. The number n = 1 corresponds to the lowest energy level, and each higher number corresponds to a higher energy level.

This table shows the relative energy of the different orbits of the Bhor hydrogen atom:

Orbit      Quantum       Energy      Relative

              number         level          energy

First          n = 1                  1                   E₁

Second    n = 2                 2                2E₁

Third        n = 3                 3                 9E₁

Fourth      n = 4                 4                16E₁

Fifth          n = 5                5                25E₁

Sixth         n = 6                6                36E₁

Seventh    n = 7                7                49E₁

When an electron jumps from a higher energy state down to a lower energy state, it emits a photon with an energy equal to the difference of the energies between the initial and the final states.

Since the n = 6 to n = 1 transition results in the higher relative energy difference (36E₁ - E₁ = 35E₁), you conclude that it is this transition which results in the emission of the highest-energy photon, which is the option A.