Question

Calculate the longest wavelength in the hydrogen emission spectrum of the Balmer series. [\(\tilde{v}=109677\ \text{cm}^{-1}\)]

• A. \(6.565\times 10^{-5}\ \text{cm}\)
• B. \(3.647\times 10^{-5}\ \text{cm}\)
• C. \(4.377\times 10^{-5}\ \text{cm}\)
• D. \(1.824\times 10^{-5}\ \text{cm}\)

Hint:

The longest wavelength of the Balmer series comes from the transition n = 3 to n = 2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Balmer lines arise when the electron falls to \(n_1=2\). The longest wavelength has the smallest energy, from \(n_2=3\).

Step 2: Rydberg Formula
\(\tilde{v}=\dfrac{1}{\lambda}=R\left(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right)=R\times\dfrac{5}{36}\)

Step 3: Calculation
\(\lambda=\dfrac{36}{5\times 109677}=6.565\times 10^{-5}\ \text{cm}\)

Step 4: Compare the options
This matches option (A).

Final Answer:
The longest Balmer wavelength is 6.565 times 10 to the power minus 5 cm. \[ \boxed{\text{(A) }6.565\times 10^{-5}\ \text{cm}} \]