Question

What is the value of frequency of radiation when transition occurs between two stationary states that differ in energy by $\Delta E$?

• A. $\nu = 2\pi h$
• B. $\nu = h\Delta E$
• C. $\nu = \frac{\Delta E}{h}$
• D. $\nu = \frac{h}{2\pi}$

Hint:

Bohr's frequency equation is a cornerstone of atomic physics: $\Delta E = h\nu$. To quickly isolate frequency, remember that energy must always be divided by Planck's constant to maintain dimensional consistency ($\text{Joules} / \text{Joules}\cdot\text{seconds} = \text{seconds}^{-1}$).
diagram diagram

Answer 1

The Correct Option is C

Step 1: Understand the question.
An electron jumps between two energy levels that differ by $\Delta E$. We must find the frequency $\nu$ of the light given out or taken in.

Step 2: Recall Bohr's frequency rule.
When an electron jumps between two levels, it releases or absorbs one photon. The photon carries energy equal to the gap between the levels.
\[ \Delta E = E_2 - E_1 \]

Step 3: Recall Planck's relation.
The energy of a photon is Planck's constant times its frequency.
\[ \Delta E = h\nu \]

Step 4: Make frequency the subject.
Divide both sides by $h$.
\[ \nu = \frac{\Delta E}{h} \]

Step 5: Check the units make sense.
Energy divided by Planck's constant gives units of per second, which is frequency. So the form is correct.

Step 6: Pick the answer.
The frequency is $\nu = \dfrac{\Delta E}{h}$, which is option 3.
\[ \boxed{\nu = \frac{\Delta E}{h}} \]