Question 387: Atoms & Nuclei

Question

In hydrogen spectrum, the ratio of wavelengths of the last line of Lyman series and that of the last line of Balmer series is

Accepted Answer

Let us find the ratio step by step.
Step 1: Formula for Wavelength in Hydrogen Spectrum
The wavelength of a spectral line in the hydrogen spectrum is given by:
$$ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) $$
For hydrogen, $Z=1$, and $R$ is the Rydberg constant.
Step 2: Lyman Series
Lyman series: $n_1 = 1$
The last line corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$.
So,
$$ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R (1 - 0) = R $$
Therefore,
$$ \lambda_L = \frac{1}{R} $$
Step 3: Balmer Series
Balmer series: $n_1 = 2$
The last line corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
So,
$$ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = R \left( \frac{1}{4} - 0 ight) = \frac{R}{4} $$
Therefore,
$$ \lambda_B = \frac{1}{R/4} = \frac{4}{R} $$
Step 4: Ratio Calculation
The required ratio:
$$ 	ext{Ratio} = \frac{\lambda_L}{\lambda_B} = \frac{\frac{1}{R}}{\frac{4}{R}} = \frac{1}{4} = 0.25 $$
Final Answer:
Option C: 0.25

Answer

1

Answer

0.5

Answer

0.25

Answer

0.2

Atoms & Nuclei

In hydrogen spectrum, the ratio of wavelengths of the last line of Lyman series and that of the last line of Balmer series is

Choose the correct answer: