Question

Given the value of Rydberg constant is $10^7 \, m^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be :

• A. $0.5 \times 10^7 \, m^{-1}$
• B. $0.25 \times 10^7 \, m^{-1}$
• C. $2.5 \times 10^7 \, m^{-1}$
• D. $0.025 \times 10^4 \, m^{-1}$

Answer 1

The Correct Option is B

To find the wave number of the last line of the Balmer series in the hydrogen spectrum, we can use the formula for the wave number (\( \bar{\nu} \)) in the hydrogen emission spectrum. The formula is given by the Rydberg formula:

\[ \bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level to which the electron falls, and \( n_2 \) is the higher energy level from which the electron falls. For the Balmer series, \( n_1 = 2 \) and \( n_2 \) should be larger.

For the last line of the Balmer series, the transition occurs from infinity to \( n=2 \). Therefore, \( n_2 = \infty \).

Using this, the formula for the wave number becomes:

\[ \bar{\nu} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{4} \right) \]

Given the Rydberg constant \( R_H = 10^7 \, \text{m}^{-1} \), substituting the values, we have:

\[ \bar{\nu} = 10^7 \times \frac{1}{4} = 0.25 \times 10^7 \, \text{m}^{-1} \]

Hence, the wave number of the last line of the Balmer series in the hydrogen spectrum is 0.25 \times 10^7 \, \text{m}^{-1}.

Conclusion: The correct option is 0.25 \times 10^7 \, \text{m}^{-1}.