Consider a microscopic spring-mass system whose spring stiffness is 39 N/m, and

Question

Consider a microscopic spring-mass system whose spring stiffness is 39 N/m, and the mass is 7.0 times 10 ^26 kg. What is the smallest amount of vibrational energy that can be added to this system? smallest energy increment = What is the difference in mass (if any) of the microscopic oscillator between being in the ground state and being in the first excited state? In a collection of these microscopic oscillators, the temperature is high enough that the ground state and the first three excited states are occupied. What are possible energies of photons emitted by these oscillators?

Explanation / Answer

(a)

E= (Ks/M)

H = 6.6e-34

= 6.6e-34 / (2) = 1.05e-34

Ks = Spring stiffness = 39 N/m

M = Mass = 7.0x10-26 kg

E = 1.05e-34 (39/7.0x10-26) = 2.4784x10-21 J

For B: You will need to calculate the change in energy (E), part C.

(highest-energy photon) - (next highest-energy photon) = lowest-energy photon

The lowest-energy photon = 2.4784x10-21

E = mc^2

E/c^2 = m

c = 3x108

2.4784x10-21/(3e8)^2 = 2.7537x10-38 kg

Energy of highest-energy photon: 2.4784x10-21 * 3 = 7.4352x10-21J
Energy of next highest-energy photon: 2.4784x10-21 * 2 =4.9568x10-21 J
Energy of lowest-energy photon: 2.4784x10-21J

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.