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(A) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2}

Vedclass · Accepted Answer

$818.9$

Vedclass · Answer

(A) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.The shortest wavelength corresponds to the transition from the highest energy level,i.e.,$n_2 = \infty$.Using the Rydberg constant $R \approx 1.097 	imes 10^7 \;m^{-1}$:$\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight)$$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes \frac{1}{9}$$\lambda = \frac{9}{1.097 	imes 10^7} \approx 8.204 	imes 10^{-7} \;m = 820.4 \;nm$.Using the energy conversion factor $13.6 \;eV = 21.76 	imes 10^{-19} \;J$ as provided in the context:$\frac{hc}{\lambda} = 21.76 	imes 10^{-19} \left( \frac{1}{3^2} - 0 ight)$$\lambda = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8 	imes 9}{21.76 	imes 10^{-19}} \approx 8.22 	imes 10^{-7} \;m = 822 \;nm$.Given the options provided,$818.9 \;nm$ is the closest value.

What is the shortest wavelength present in the Paschen series of spectral lines (in $;nm$)?

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