Question

Review Constants Let's use the Bohr model equations to explore some properties of the hydrogen atom. We will determine the kinetic, potential, and total energies of the hydrogen atom in the n=2 state, and find the wavelength of the photon emitted in the transition n = 2 →n=1. En= 1 med 602 8nºh? -13.6 eV 722 These two equations are equal only if the coefficients of 1/n2 are equal: met =hcR= 13.6 eV 8€0h? SOLVE Using this equation, we can express the kinetic, potential, and total energies for the state with quantum number n, respectively, as K = 13.6 eV -27.2 eV Un = - 13.6eV E = 72 724 For the n=2 state, we find that K2 = 3.40 eV, U2 = -6.80 eV, and E2 = -3.40 eV. The energy of the emitted photon is E - Ef = -3.40 eV - (-13.6 eV) = 10.2 eV 1.63 x 10-18 J This is equal to hc/X, so we obtain = = hc (6.626 x 10-31 J .s)(3.00 x 108 m/s) E-E 1.63 x 10-18 122 x 10-9 m= 122 nm REFLECT The wavelength we just found is that of the Lyman alpha line, the longest-wavelength line in the Lyman series of ultraviolet lines in the hydrogen spectrum. Part A - Practice Problem: Find the wavelength for the transition n=4 →n=1 for singly ionized helium, which has one electron and a nuclear charge of 2e. (Note that the value of the Rydberg constant is four times as great as for hydrogen because it is proportional to the square of the product of the nuclear charge and the electron charge.) Express your answer in nanometers to three significant figures. EVO AXO ? nm Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining

answer is not 122 or 97.2

Answer 1

1/$\lambda$ = 4R ( 1/nf² - 1/ni²)

1/$\lambda$ = 4 * 1.097e7 ( 1/1² - 1/4²)

1/$\lambda$ = 41137500

$\lambda$ = 2.431e-8 m

or

$\lambda$ = 24.3 nm