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(B) The wavelength $\lambda$ of emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \ri

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$n=5$ to $n=3$

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(B) The wavelength $\lambda$ of emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Ultraviolet radiation corresponds to the Lyman series $(n_1 = 1)$.Infrared radiation corresponds to the Paschen series $(n_1 = 3)$,Brackett series $(n_1 = 4)$,or Pfund series $(n_1 = 5)$.Comparing the options:$(A)$ $n=6$ to $n=2$ is the Balmer series (Visible).$(B)$ $n=5$ to $n=3$ is the Paschen series (Infrared).$(C)$ $n=3$ to $n=5$ is an absorption process,not emission.$(D)$ $n=4$ to $n=2$ is the Balmer series (Visible).Therefore,the transition $n=5$ to $n=3$ results in infrared radiation.

In hydrogen atoms,the transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

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