Question

For wavelength of visible radiation of Hydrogen spectrum, Balmer gave an equation as \(λ = \frac{xm^2}{m^2-4}\), where m is the integer value. The value of x in terms of Rydberg's constant R is

• A. \(\frac{R}{4}\)
• B. \(\frac{4}{R}\)
• C. \(2R\)
• D. \(4R\)

Hint:

Start from \(\dfrac1\lambda=R\left(\dfrac1{2^2}-\dfrac1{m^2}\right)\) and invert.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The Balmer series arises from transitions to \(n=2\) in hydrogen: \(\dfrac1\lambda=R\left(\dfrac14-\dfrac1{m^2}\right)\).

Step 2: Key Formula or Approach
Combine the fractions in the bracket.

Step 3: Detailed Explanation
\[ \frac1\lambda=R\,\frac{m^2-4}{4m^2} \]
\[ \lambda=\frac{4}{R}\cdot\frac{m^2}{m^2-4} \]
Comparing with \(\lambda=\dfrac{xm^2}{m^2-4}\) gives \(x=\dfrac4R\).

Final Answer:
The constant is \(x=\frac4R\), option (B). \[ \boxed{\dfrac4R\ \text{(B)}} \]