When an electron in a hydrogen atom jumps from n = 2 to n = 1, the photon emitted is of frequency f = 4.572 $\times$ 10¹⁴ Hz. What is the energy of the photon? A) 2.187 $\times$ 10^-15 J B) 3.029 $\times$ 10^-19 J C) 3.33 $\times$ 10^-9 J D) 6.623 $\times$ 10^-34 J E) None of these is correct.

Question

Quizplus · Accepted Answer

The energy of the photon emitted can be calculated using the formula E=hf, where E is energy, h is Planck's constant, and f is frequency. 
Using the given frequency, we get: 
E = (6.626 x 10^-34 J s) x (4.572 x 10^14 Hz) = 3.029 x 10^-19 J 
Converting to electronvolts (eV): 
E = 3.029 x 10^-19 J / 1.602 x 10^-19 J/eV = 1.89 eV 
Therefore, the energy of the photon emitted is 1.89 eV, which corresponds to choice B in Joules.

When an Electron in a Hydrogen Atom Jumps from N ×\times×

When an Electron in a Hydrogen Atom Jumps from N $\times$