Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$ . The shortest wavelength in the Bracket series is

$2\lambda$

$4\lambda$

$9\lambda$

$16\lambda$

Step-by-Step Solution

Shortest wavelength $\frac{1}{\lambda} = R(\frac{1}{n_f^2} - 0)$ . For Balmer, $n_f=2$ , $\frac{1}{\lambda} = R(\frac{1}{4})$ . For Bracket, $n_f=4$ , $\frac{1}{\lambda'} = R(\frac{1}{16})$ . Thus $\frac{\lambda'}{\lambda} = \frac{16}{4} = 4$ , so $\lambda' = 4\lambda$ .

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