Question

Calculate the shortest wavelength in hydrogen spectrum emission of Lymen series (\(R_H = 109677\,\text{cm}^{-1}\))

• A. \(9.117\times 10^{-6}\) cm
• B. \(9.241\times 10^{-6}\) cm
• C. \(9.360\times 10^{-6}\) cm
• D. \(9.482\times 10^{-6}\) cm

Hint:

For the series limit of Lyman, put n1 = 1 and n2 = infinity in the Rydberg formula.

Answer 1

The Correct Option is A

Step 1: Understand the question
The shortest wavelength of a spectral series is the series limit. It comes from a jump from \(n_2 = \infty\) down to the lowest level of that series. For the Lyman series the lowest level is \(n_1 = 1\).

Step 2: Write the Rydberg formula
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Step 3: Substitute the limits
With \(n_1 = 1\) and \(n_2 = \infty\), the second term vanishes.
\[ \frac{1}{\lambda} = 109677 \times (1 - 0) = 109677 \ \text{cm}^{-1} \]

Step 4: Find the wavelength
\[ \lambda = \frac{1}{109677} = 9.117 \times 10^{-6} \ \text{cm} \]

Step 5: Check the other options
The other values are slightly longer than the true limit. A transition from a finite level such as \(n_2 = 6\) gives a longer wavelength than the limit, and no transition can give a wavelength below \(1/R_H\). So only 9.117e-6 cm is the series limit.

Final Answer:
The Lyman series limit is 9.117 x 10^-6 cm. This is option (A). \[ \boxed{\text{(A) }9.117 \times 10^{-6} \ \text{cm}} \]