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Question: What will be the longest wavelength line in the Balmer series of spectrum of H-atom. A. \(546nm\) ...

What will be the longest wavelength line in the Balmer series of spectrum of H-atom.
A. 546nm546nm
B. 656nm656nm
C. 566nm566nm
D. 556nm556nm

Explanation

Solution

In the Balmer series, an electron moves from a higher energy orbital to the second orbital. The energy emitted by electron transitions can be calculated using the formula E=hcλE = \dfrac{{hc}}{\lambda } . In the Blamer series, we'll employ the energy released equation. To determine the longest wavelength, the amount of energy produced must be kept to a minimum.

Formula used:
1λ=RZ2(1n12−1n22)\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)

Complete step by step solution:
The Balmer series is a section of the hydrogen electromagnetic spectrum with a distinct starting and ending orbital through which the electronic transition occurs.
Now let's look at the electrical transition between two orbits that occurs when an electron is liberated as a spectrum. Whatever orbital an electron jumps from, the Balmer series corresponds to transitions from a higher orbital to the second innermost orbital.
In the case of single electron species,
1λ=RZ2(1n12−1n22)\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)
Where RR denotes the Rydberg constant =1.097×107×10−9nm−1=1.097×10−2nm−1 = 1.097 \times {10^7} \times {10^{ - 9}}n{m^{ - 1}} = 1.097 \times {10^{ - 2}}n{m^{ - 1}}
Z = Atomic number = 1 for the hydrogen atom
Because wavelength is inversely related to energy, the lowest energy transition should be addressed for the longest wavelength in the Balmer series n1=2{n_1} = 2
As a result, n2=3{n_2} = 3
∴1λ=1.097×10−2nm−1×12(122−132) ⇒λ=656nm  \therefore \dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}n{m^{ - 1}} \times {1^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) \\\ \Rightarrow \lambda = 656nm \\\
Therefore, the longest wavelength line in Balmer series of spectrum of H-atom is λ=656nm\lambda = 656nm
So, the correct option is: B. 656nm656nm

Note: Students should recall the meaning of the Balmer series and not be confused by orbital numbers connected to the transition in the Balmer series; instead, students should estimate the final orbital while paying attention to the signs and relationships in the formula.