Energy levels for mass confined to 1D box

AI Thread Summary
For a nitrogen molecule confined in a 1D box, the lowest two energy levels were calculated using the formula E = n².h²/8mL², resulting in E1 = 2.358x10^-24 J and E2 = 9.432x10^-42 J. The characteristic temperature (Tc) was determined to be approximately 0.512 Kelvin, indicating that the system remains in the ground state only at very low temperatures. Concerns were raised about the reasonableness of this temperature, given the unfamiliarity with characteristic temperature in such contexts. The calculations were confirmed to be correct, suggesting that thermal excitation would likely place the system in excited states at higher temperatures.
SalfordPhysics
Messages
68
Reaction score
1

Homework Statement


For a nitrogen molecule, calculate the lowest 2 energy levels and the characteristic temperature;
Mass of molecule = 2.33x10-26[kg]
Length of box = 10-9[m]

Homework Equations


E = n2.h2/8mL2 (n=1,2,3,...)

Characteristic Temperature (Tc) -> when thermal energy kBT = energy spacing of lowest 2 levels.

The Attempt at a Solution


[/B]
E1 = h2/(8).(2.33x10-44)
= 2.358x10-24 [J]

E2 = n2.(E1) = 4(E1) = 9.432x10-42

Tc = (E2 - E1)/kB = 7.074x10-24 / kb = 0.512 [Kelvin]

The final answer quite simply does;t seem right to me, but I have checked numerous times my calculations. Can anyone clarify please.
 
Physics news on Phys.org
Why does it sound wrong?
 
Just doesn't seem like 0.5 Kelvins is a reasonable answer. Main problem is that the term characteristic temperature is a new one and not quite sure of its understanding.
 
Unless you have some common experience with nanometer size boxes, how can you even expect that you can tell what is a reasonable answer?
Your calculations seem OK. The result means that the system will be in the ground state only at very low temperatures. Otherwise will be very likely in excited states, due to the thermal excitation.
 
  • Like
Likes SalfordPhysics
Thanks nasu, this makes sense.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top