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QJEEM23Ch627a: JEE Main 2023 Chemistry - Class 11 JEEM23Ch627a Solution: If wavelength of the first line of the Paschen series of…

If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _________ nm. (Nearest integer)

Step-by-Step Verified Solution

1720=R×(19−116)\frac{1}{720}=R \times\left(\frac{1}{9}-\frac{1}{16}\right)

amp;⇒R=9×16720×7amp;1λ′=9×16720×7×(19−125)amp;λ′=492.18 nmamp;λ′=492 nm (nearest integer) \begin{aligned}& \Rightarrow R=\frac{9 \times 16}{720 \times 7} \\\\& \frac{1}{\lambda^{\prime}}=\frac{9 \times 16}{720 \times 7} \times\left(\frac{1}{9}-\frac{1}{25}\right) \\\\& \lambda^{\prime}=492.18 \mathrm{~nm} \\\\& \lambda^{\prime}=492 \mathrm{~nm} \text { (nearest integer) }\end{aligned}

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