An electron in an atom drops from an energy level at -1.1 *10^-18 J to an energy level at -2.4* 10^-18 J.The wave associated with the emitted photon has a frequency of: A)2.0 * 10¹⁷ Hz B)2.0 *10¹⁵ Hz C)2.0 * 10¹³ Hz D)2.0 * 10¹¹ Hz E)2.0 *10⁹ Hz

Question

Quizplus · Accepted Answer

The frequency of the emitted photon can be calculated using the formula: ΔE = hf where ΔE is the energy difference between the initial and final levels, h is Planck's constant and f is the frequency of the emitted photon. Thus, f = ΔE/h.

Substituting the given values, we get:

ΔE = (-2.4 x 10^-18 J) - (-1.1 x 10^-18 J) = -1.3 x 10^-18 J

h = 6.626 x 10^-34 J s

f = (-1.3 x 10^-18 J) / (6.626 x 10^-34 J s) = 1.963 x 10^15 Hz ≈ 2.0 x 10^15 Hz

Therefore, the best choice is B) 2.0 x 10^15 Hz.

An Electron in an Atom Drops from an Energy Level