The first member of the Balmer series of hydrogen atom has a wavelength of 6561 ∘ A . The wavelength of the second member of the Balmer series (in nm) is_____________

Correct answer: 0 (1)/(λ) = R Z ^ 2 ( 1 n _ 1 ^ 2 - 1 n _ 2 ^ 2 ) 1 λ _ 1 = R (1) ^ 2 ( 1 2 ^ 2 - 1 4 ^ 2 ) = (3R)/(16) λ _ 2 λ _ 1 = (20)/(27) λ _ 2 = (20)/(27) × 6561 ∘ A = 4860 ∘ A = 486nm

JEE Main 20208 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Evening Shift
Subject: Physics
Chapter: Atomic Physics

JEE Main 2020 · 8 January · Evening Shift · Physics · Atomic Physics

The first member of the Balmer series of hydrogen atom has a...

The first member of the Balmer series of hydrogen atom has a wavelength of

6561\overset{∘}{A}

. The wavelength of the second member of the Balmer series (in nm) is_____________

Correct answer

Step-by-step solution

\dfrac{1}{\lambda} = R{Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right)

\dfrac{1}{{\lambda}_{1}} = R{\left(1\right)}^{2}\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{4}^{2}}\right) = \dfrac{3R}{16}

\dfrac{{\lambda}_{2}}{{\lambda}_{1}} = \dfrac{20}{27}

{\lambda}_{2} = \dfrac{20}{27} \times 6561\overset{∘}{A} = 4860\overset{∘}{A}

= 486nm

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