A mass of 0.250 kg is oscillating with small amplitude oscillations on the end of a spring whose stiffness constant is 10 N/m.When this oscillator makes a transition from its n = 6 state to its n = 5 state,the energy of the photon emitted is A)6.67 $\times$ 10^-34 J B)1.33 $\times$ 10^-33 J C)3.33 $\times$ 10^-34 J D)1.00 $\times$ 10^-33 J E)None of these is correct.

Question

Quizplus · Accepted Answer

The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the photon. The frequency of a photon emitted when an oscillator makes a transition from state n to state n-1 is given by f = (E_n - E_(n-1))/h, where E_n is the energy of the oscillator in state n. The energy of an oscillator in state n is given by E_n = (n + 1/2)hf_0, where f_0 is the natural frequency of the oscillator. Substituting these expressions into the equation for the frequency of a photon, we get f = (n + 1/2)f_0 - (n - 1/2)f_0 = f_0. Therefore, the energy of a photon emitted when an oscillator makes a transition from state n to state n-1 is given by E = hf = hf_0. Substituting the given values into this equation, we get E = (6.626 x 10^-34 J s)(10 Hz) = 6.626 x 10^-33 J.

A Mass of 0 ×\times× 10-34 J B)1 ×\times× 10-33 J C)3 ×\times×

A Mass of 0 $\times$ 10^-34 J
B)1 $\times$ 10^-33 J
C)3 $\times$