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QJEEM20Phbfc7: JEE Main 2020 Physics - Class 12 JEEM20Phbfc7 Solution: 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 Å. The wavelength of the second member of the Balmer series (in nm) is:

Step-by-Step Verified Solution

1λ1=RZ2(122−132){1 \over {{\lambda _1}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = 536{5 \over {36}} RZ2

1λ2=RZ2(122−142){1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = 1264{{12} \over {64}} RZ2

∴\therefore λ2λ1{{{\lambda _2}} \over {{\lambda _1}}} = 536×6412{5 \over {36}} \times {{64} \over {12}} = 2027{{20} \over {27}}

⇒\Rightarrow λ\lambda 2 = 2027λ1{{20} \over {27}}{\lambda _1}

= 2027{{20} \over {27}} ×\times 6561 = 4860 Ao\mathop A\limits^o = 486 nm

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