Q.A difference of $2.3\ \text{eV}$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

The key idea is that the energy of the emitted photon equals the energy difference between the two levels, given by $E = h f$.

Step 1: Write the relation between energy and frequency:

where $h = 4.135667696 \times 10^{-15}\ \text{eV·s}$ (Planck's constant in eV·s).

Step 2: The energy difference is $E = 2.3\ \text{eV}$. Solve for $f$:

$f = \frac{E}{h} = \frac{2.3}{4.135667696 \times 10^{-15}}$ …

When an electron in an atom jumps from a higher energy level to a lower one, the atom loses energy. That energy doesn't vanish — it leaves the atom as a single packet of light, a photon. The photon’s energy is exactly equal to the difference between the two atomic energy levels.

This is the core idea behind atomic spectra: each spectral line corresponds to a specific transition, and the photon’s frequency is locked to the energy gap by Planck’s constant $h$.

Here $h = 6.626 \times 10^{-34}\ \text{J·s}$ is Planck’s constant, and $\nu$ is the frequency in hertz. The problem gives $\Delta E = 2.3\ \text{eV}$, but $h$ is in joules — so we must convert the energy to joules first.

1. Convert the energy from eV to joules. The conversion factor is $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$. So:

2. Apply the photon energy relation. From $E = h\nu$, we solve for frequency:

3. Plug in the numbers.

Divide:

$\nu = 5.56 \times 10^{14}\ \text{Hz}$ …

Method: Energy–Frequency Relation for a Photon (Planck–Einstein Relation)

This is a direct application of the Planck–Einstein equation, which connects the energy of a photon to its frequency. When an electron drops from a higher energy level to a lower one, the energy lost is emitted as a single photon. The photon’s energy equals the difference between the two atomic energy levels.

Step 1 – Write the Planck–Einstein relation

The energy $E$ of a photon is proportional to its frequency $\nu$:

where $h$ is Planck’s constant. In SI units, $h = 6.626 \times 10^{-34}\ \text{J·s}$. But here the energy is given in electronvolts (eV), so we need the value of $h$ in eV·s:

(You may use $4.14 \times 10^{-15}\ \text{eV·s}$ in exams unless more precision is specified.)

Step 2 – Identify the photon energy

The energy difference between the two levels is given as $\Delta E = 2.3\ \text{eV}$. This entire amount is carried away by the emitted photon:

Step 3 – Solve for frequency

Rearrange $E = h\nu$ to get:

Substitute the numbers: …

Students often handle this question well because it is a direct application of the formula $E = h\nu$. But a few predictable mistakes trip people up. Here are the most common ones and how to avoid each.

1. Forgetting to convert electron-volts to joules

The energy difference is given in eV, but Planck’s constant $h$ is usually taken in SI units ($6.63 \times 10^{-34}\ \text{J·s}$). If you plug $2.3\ \text{eV}$ directly into $E = h\nu$, you get a completely wrong frequency.

How to avoid: Always convert eV to joules before using $h$ in J·s.

$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$, so

$E = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19}\ \text{J}$.

2. Using the wrong formula (confusing frequency with wavelength)

Some students reach for $E = \frac{hc}{\lambda}$ instead of $E = h\nu$. The question asks for frequency, not wavelength. Using the wrong formula wastes time and often leads to an incorrect unit (metres instead of Hz).

How to avoid: Read the question carefully. If it says “frequency”, think $\nu = \frac{E}{h}$. If it says “wavelength”, think $\lambda = \frac{hc}{E}$. There is no shortcut — you must match the formula to the quantity asked.

3. Arithmetic errors with powers of ten

When you divide $3.68 \times 10^{-19}$ by $6.63 \times 10^{-34}$, the exponents are $(-19) - (-34) = +15$. Students sometimes subtract incorrectly and get $10^{-53}$ or $10^{53}$.

How to avoid: Write the division step clearly: …