Question

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to orbit of level $n=2$, then the emitted radiation frequency is (where $R=$ Rydberg's constant, $C=$ Velocity of light)

• A. $\frac{3RC}{27}$
• B. $\frac{RC}{25}$
• C. $\frac{8RC}{9}$
• D. $\frac{5RC}{36}$

Hint:

Logic Tip: This specific transition from $n=3$ to $n=2$ produces the first line of the Balmer series in the hydrogen spectrum.

Answer 1

The Correct Option is D

Concept:
When an electron jumps from a higher energy level to a lower energy level in a hydrogen atom, radiation is emitted. The wavelength of emitted radiation is given by the Rydberg formula: \[ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] where:

• $R$ = Rydberg constant
• $n_1$ = lower energy level
• $n_2$ = higher energy level
• $n_2>n_1$

Frequency is related to wavelength by: \[ f=\frac{c}{\lambda} \] where $c$ is speed of light.
Step 1: Given transition
Electron jumps from: \[ n_2=3 \quad \text{to} \quad n_1=2 \] This is a Balmer series transition.
Step 2: Apply Rydberg formula
\[ \frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \] \[ \frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{9}\right) \] Take LCM = 36: \[ \frac{1}{\lambda}=R\left(\frac{9-4}{36}\right) \] \[ \frac{1}{\lambda}=R\left(\frac{5}{36}\right) \] \[ \frac{1}{\lambda}=\frac{5R}{36} \]
Step 3: Find frequency
We know: \[ f=\frac{c}{\lambda} \] Since: \[ \frac{1}{\lambda}=\frac{5R}{36} \] Therefore: \[ f=c\left(\frac{1}{\lambda}\right) \] \[ f=c\left(\frac{5R}{36}\right) \] \[ f=\frac{5Rc}{36} \]
Step 4: Final Answer
The frequency of emitted radiation is: \[ \boxed{\frac{5Rc}{36 \] Quick Tip:
Always first calculate $\frac{1}{\lambda}$ using Rydberg formula, then use $f=c/\lambda$.