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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)
1720=R ×(19−116)\frac{1}{720}=R \times\left(\frac{1}{9}-\frac{1}{16}\right)7201=R ×(91−161)
⇒ R=9 × 16720 × 7 1λ′=9 × 16720 × 7 ×(19−125) λ′=492.18 nm 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} { (nearest integer) } \end{aligned} ⇒ R=720 × 79 × 16 λ′1=720 × 79 × 16 ×(91−251) λ′=492.18 nm λ′=492 nm (nearest integer)
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