Wavelength of Photon Emitted During n = 3 to n = 2 Transition in Hydrogen Atom

Understanding the Electron Transition in the Hydrogen Atom

This question is based on the quantized energy levels of the hydrogen atom. An electron in a hydrogen atom can occupy only certain allowed energy states, represented by the principal quantum number n. When an electron moves from a higher energy level to a lower energy level, the difference in energy between the two states is released in the form of a photon.

Here, the electron makes a transition from the n = 3 energy level to the n = 2 energy level. Since the electron moves from a higher energy state to a lower energy state, energy is emitted rather than absorbed. The emitted photon has an energy exactly equal to the difference between the energies of the n = 3 and n = 2 states.

To determine the wavelength of this photon, we first calculate the energy released during the electronic transition. We then use the relationship between photon energy and wavelength to obtain the required answer in nanometres.

Energy Levels of the Hydrogen Atom

The energy of an electron in the n^th energy level of a hydrogen atom is given by:

E_n = −R_H/n²

Here, E_n is the energy of the electron in the n^th state, R_H is the Rydberg energy constant, and n is the principal quantum number.

The negative sign indicates that the electron is bound to the nucleus. An electron in a higher value of n has a less negative energy and is therefore in a higher energy state.

Step 1: Calculate the Energy Released During the Transition

For a transition from an initial state n_i to a lower final state n_f, the magnitude of the energy emitted as a photon is:

ΔE = R_H[(1/n_f²) − (1/n_i²)]

In the present question:

n_i = 3

n_f = 2

Therefore:

ΔE = R_H[(1/2²) − (1/3²)]

Substituting R_H = 2.18 × 10⁻¹⁸ J:

ΔE = 2.18 × 10⁻¹⁸ [(1/4) − (1/9)]

Taking the difference of the two fractions:

1/4 − 1/9 = (9 − 4)/36 = 5/36

Therefore:

ΔE = 2.18 × 10⁻¹⁸ × 5/36

ΔE ≈ 3.03 × 10⁻¹⁹ J

Thus, the energy of the photon emitted during the n = 3 to n = 2 transition is approximately 3.03 × 10⁻¹⁹ J.

Step 2: Relate Photon Energy to Wavelength

The energy of a photon is related to its wavelength by the equation:

E = hc/λ

Here, E is the energy of the photon, h is Planck’s constant, c is the speed of light, and λ is the wavelength of the emitted photon.

Rearranging the equation to calculate the wavelength:

λ = hc/E

For this transition, E = ΔE. Therefore:

λ = hc/ΔE

Step 3: Substitute the Given Values

Substituting h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, and ΔE ≈ 3.03 × 10⁻¹⁹ J:

λ = [(6.626 × 10⁻³⁴) × (3 × 10⁸)]/(3.03 × 10⁻¹⁹)

Multiplying the values in the numerator:

hc = 19.878 × 10⁻²⁶ J m

hc = 1.9878 × 10⁻²⁵ J m

Therefore:

λ = (1.9878 × 10⁻²⁵)/(3.03 × 10⁻¹⁹)

λ ≈ 6.57 × 10⁻⁷ m

Step 4: Convert the Wavelength From Metres to Nanometres

The question asks for the wavelength in nanometres. We know that:

1 nm = 10⁻⁹ m

Therefore:

λ = (6.57 × 10⁻⁷)/(10⁻⁹) nm

λ ≈ 6.57 × 10² nm

λ ≈ 656.5 nm

Since the question asks for the answer as an integer, the value is rounded to the nearest whole number:

λ ≈ 657 nm

Direct Calculation Using the Combined Formula

The wavelength can also be calculated directly by combining the hydrogen energy-level equation with the photon energy relation:

λ = hc/{R_H[(1/n_f²) − (1/n_i²)]}

For the transition from n = 3 to n = 2:

λ = [(6.626 × 10⁻³⁴) × (3 × 10⁸)]/{(2.18 × 10⁻¹⁸)[(1/4) − (1/9)]}

Using:

(1/4) − (1/9) = 5/36

We obtain:

λ ≈ 6.565 × 10⁻⁷ m

Therefore:

λ ≈ 656.5 nm ≈ 657 nm

Why Is a Photon Emitted in the n = 3 to n = 2 Transition?

The n = 3 state has a higher energy than the n = 2 state. When the electron moves from n = 3 to n = 2, it loses energy. According to the law of conservation of energy, this lost energy cannot disappear and is released as electromagnetic radiation in the form of a photon.

The energy of the emitted photon is exactly equal to the difference between the two atomic energy levels. A larger energy difference would produce a photon of shorter wavelength, while a smaller energy difference would produce a photon of longer wavelength.

Connection With the Balmer Series of Hydrogen Spectrum

The transition from n = 3 to n = 2 belongs to the Balmer series of the hydrogen emission spectrum. The Balmer series consists of all electronic transitions from higher energy levels to the n = 2 level.

The n = 3 to n = 2 transition produces the well-known H-alpha spectral line. Its wavelength is approximately 656 nm, placing it in the red region of the visible spectrum. The small difference between the commonly quoted value and the calculated value here arises from the rounded physical constants provided in the question.

Physical Interpretation of the Calculated Wavelength

The calculated wavelength of approximately 657 nm lies in the visible region of the electromagnetic spectrum. This means that the photon emitted during the n = 3 to n = 2 transition can be observed as red light.

This characteristic wavelength is one of the most important spectral lines of hydrogen. Because every hydrogen atom has the same quantized energy levels, the same electronic transition always produces a photon with the same characteristic wavelength under ideal conditions.

Final Answer

Wavelength of the emitted photon = 657 nm

For the transition from n = 3 to n = 2 in a hydrogen atom, the energy released is calculated using the Rydberg energy relation. The emitted photon has a wavelength of approximately 656.5 nm. Therefore, when expressed as an integer, the required answer is 657 nm.