The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to (use R_H=1.097 10^7 m^-1 )

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Accepted Answer

Correct answer: A. (a) For shortest wavelength. n_f=2 and n_i= So, 1 _ short=R_H( 1n_f^2- 1n_i^2) ⇒ 1 _s=1.097 10^7( 12^2- 1 ^2) ⇒ 1 _s=2742500 m^-1 ⇒ _s=3646 10^-10 m

$The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to (use R_{H} = 1.097 \times 10^{7} m^{- 1})$

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