Question

The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies $E_1$ (lower) and $E_2$ (higher) is given by:

• A. $v=\frac{E_1+E_2}{h}$
• B. $v=\frac{E_1-E_2}{h}$
• C. $v=\frac{E_1\times E_2}{h}$
• D. $v=\frac{E_2-E_1}{h}$

Answer 1

Answer: (D)

$v=\frac{E_2-E_1}{h}$

According to Bohr's postulates:

The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by $\Delta E$, is given by:

$\nu =\frac{\Delta E}{h}=\frac{E_2-E_1}{h}$

Where $E_1$ and $E_2$ are the energies of the lower and higher allowed energy states, respectively.

This expression is commonly known as Bohr’s frequency rule.