Question

The difference between the frequencies of the first and second Lyman lines of hydrogen atom is (R - Rydberg constant and c - speed of light in vacuum)

• A. $\dfrac{9Rc}{28}$
• B. $\dfrac{7Rc}{12}$
• C. $\dfrac{3Rc}{8}$
• D. $\dfrac{5Rc}{36}$

Hint:

Remember to apply the Rydberg formula carefully and subtract frequencies to find line differences.

Answer 1

The Correct Option is D

The frequency of the Lyman series transition from level $n$ to 1 is given by the Rydberg formula:
\[ \nu = Rc \left(1 - \frac{1}{n^2}\right) \] - For the first Lyman line ($n=2 \to 1$):
\[ \nu_1 = Rc \left(1 - \frac{1}{2^2}\right) = Rc \left(1 - \frac{1}{4}\right) = \frac{3Rc}{4} \] - For the second Lyman line ($n=3 \to 1$):
\[ \nu_2 = Rc \left(1 - \frac{1}{3^2}\right) = Rc \left(1 - \frac{1}{9}\right) = \frac{8Rc}{9} \] The difference between the frequencies is:
\[ \Delta \nu = \nu_2 - \nu_1 = \frac{8Rc}{9} - \frac{3Rc}{4} = Rc \left(\frac{8}{9} - \frac{3}{4}\right) = Rc \left(\frac{32 - 27}{36}\right) = \frac{5Rc}{36} \] Hence, the difference is $\dfrac{5Rc}{36}$.