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: Understanding the Question:
The question asks for the mathematical expression of the frequency of emitted or absorbed electromagnetic radiation when an atomic electron undergoes a transition between two energy levels separated by an energy difference of $\Delta E$.

Step 2: Detailed Explanation:
According to Bohr's frequency rule (derived from Max Planck's quantum theory), when an electron transitions between two stationary energy states, energy is emitted or absorbed in the form of a single discrete photon.
The energy of this photon is exactly equal to the difference in potential energies between the two stationary orbits:
$$ \Delta E = E_2 - E_1 $$ The energy associated with a single quantum photon is directly proportional to its operational frequency ($\nu$), governed by the Planck relation:
$$ \Delta E = h\nu $$ Here, $h$ represents Planck's constant. Isolating the frequency parameter ($\nu$) from this equation yields:
$$ \nu = \frac{\Delta E}{h} $$

Step 3: Final Answer:
The frequency expression is $\nu = \frac{\Delta E}{h}$, which directly matches option (C).