In the formula 1/$\lambda$ = R (1 / n_f² - 1/n_i²), Balmer found that for the visible lines in hydrogen that n_f = _______. A) 1 B) 2 C) 3 D) 4 E) 5

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Balmer's formula relates the wavelengths of the visible lines in the hydrogen atomic emission spectrum to the energy levels of the hydrogen atom. The formula is given as: 1/λ = R(1/n_f^2 - 1/n_i^2) where λ is the wavelength of the emitted light, R is the Rydberg constant, and n_f and n_i are the final and initial energy levels of the atom, respectively. For the visible lines in hydrogen, we know that the final energy level is n_f = 2, since those are the transitions that produce visible light. Plugging this into the formula and solving for n_i, we get: 1/λ = R(1/2^2 - 1/n_i^2) 1/n_i^2 = 1/(R(1/λ) + 1/2^2) n_i = sqrt(1/(R(1/λ) + 1/2^2)) This means that n_f = 2, and we can eliminate choices A, C, D, and E. The correct answer is B.

In the Formula 1 λ\lambdaλ = R (1 / Nf2 - 1/ni2), Balmer Found That

In the Formula 1 $\lambda$ = R (1 / N_f² - 1/n_i²), Balmer Found That