The first member of the Balmer series of hydrogen atom has a wavelength of $6561 \AA$. The wavelength of the second member of the Balmer series (in nm) is ____.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Modern Physics

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 ____.

Answer: 486

Solution

For the Balmer series,

\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{n ^{2}})

. The first member (

n = 3

) gives

\frac{1}{λ _{1}} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}

and the second (

n = 4

) gives

\frac{1}{λ _{2}} = R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R}{16}

. Dividing,

\frac{λ _{2}}{λ _{1}} = \frac{5/36}{3/16} = \frac{20}{27}

, so

λ_{2} = \frac{20}{27} \times 6561 = 4860

Å, which is

486

nm.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer

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