Calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level.Recall that for hydrogen En = -2.18 × 10^-1⁸ J(1/n² ).(h = 6.63 × 10^-34 J • s) A)3.08 × 10¹⁵ s^-1 B)1.03 × 10⁸ s^-1 C)2.06 × 10¹⁴ ^s-1 D)1.35 × 10^-51 ^s-1 E)8.22 × 10¹⁴ ^s-1

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The correct answer is A.The frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level can be calculated using the formula:$$f = \frac{\Delta E}{h}$$where:* f is the frequency of the light emitted* ΔE is the energy difference between the two energy levels* h is Planck's constantThe energy difference between the two energy levels can be calculated using the formula:$$\Delta E = E_f - E_i$$where:* Ef is the energy of the final energy level* Ei is the energy of the initial energy levelThe energy of a hydrogen atom in a particular energy level is given by the formula:$$E_n = -\frac{2.18 	imes 10^{-18} J}{n^2}$$where:* En is the energy of the atom in the nth energy level* n is the principal energy levelSubstituting the values of Ef and Ei into the formula for ΔE, we get:$$\Delta E = -\frac{2.18 	imes 10^{-18} J}{1^2} - \left(-\frac{2.18 	imes 10^{-18} J}{4^2}ight)$$$$\Delta E = -2.18 	imes 10^{-18} J + \frac{2.18 	imes 10^{-18} J}{16}$$$$\Delta E = -2.18 	imes 10^{-18} J + 0.13625 	imes 10^{-18} J$$$$\Delta E = -2.04375 	imes 10^{-18} J$$Substituting the values of ΔE and h into the formula for f, we get:$$f = \frac{-2.04375 	imes 10^{-18} J}{6.63 	imes 10^{-34} J \cdot s}$$$$f = 3.08 	imes 10^{15} s^{-1}$$Therefore, the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is 3.08 × 10^15 s^-1.

Calculate the Frequency of the Light Emitted by a Hydrogen