Question 435: Atoms & Nuclei

Question

The ratio of the wavelength of the last line of Paschen series to that of Balmer series is

Accepted Answer

Step 1: Formula for wavelength in hydrogen spectrum (Rydberg formula)
The wavelength of a spectral line is given by:
$$ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) $$
where:
$R$ = Rydberg constant
$Z$ = atomic number (for hydrogen, $Z=1$)
$n_1$ = lower energy level
$n_2$ = higher energy level, $n_2 > n_1$
Step 2: Balmer series "last line"
Balmer series: $n_1 = 2$, $n_2 = \infty$ (last line means limit, $n_2 	o \infty$):
So, for the last line:
$$ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 ight) = R \cdot \frac{1}{4} $$
So,
$$ \lambda_B = \frac{1}{R} 	imes 4 = \frac{4}{R} $$
Step 3: Paschen series "last line"
Paschen series: $n_1 = 3$, $n_2 = \infty$ (last line):
So,
$$ \frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - 0 ight) = R \cdot \frac{1}{9} $$
So,
$$ \lambda_P = \frac{1}{R} 	imes 9 = \frac{9}{R} $$
Step 4: Ratio of wavelengths
$$ 	ext{Ratio} = \frac{\lambda_P}{\lambda_B} = \frac{ \frac{9}{R} }{ \frac{4}{R} } = \frac{9}{4} $$
Step 5: Final Answer
Correct option: A $\boxed{\frac{9}{4}}$

Answer

$\frac{9}{4}$

Answer

$\frac{3}{2}$

Answer

$\frac{2}{3}$

Answer

$\frac{4}{9}$

Atoms & Nuclei

The ratio of the wavelength of the last line of Paschen series to that of Balmer series is

Choose the correct answer: