An electron is trapped in a square well of unknown width, L. It starts in unknown energy level, n. When it falls to level n-1 it
emits a photon of wavelength λphoton = 2280 nm. When it falls from n-1 to n-2, it emits a photon of wavelength λphoton = 3192 nm.
1) What is the energy of the n to n-1 photon in eV?
En to n-1 =
2) What is the energy of the n-1 to n-2 photon in eV?
En-1 to n-2 =
3) What is the initial value of n?
ninitial =
4) What is the width, L, of the well in nm?
L =
5) What is the longest wavelength of light, λlongest, the well can absorb in nm?
λlongest =
1) What is the energy of the n to n-1 photon in eV?
En to n-1 =
2) What is the energy of the n-1 to n-2 photon in eV?
En-1 to n-2 =
3) What is the initial value of n?
ninitial =
4) What is the width, L, of the well in nm?
L =
5) What is the longest wavelength of light, λlongest, the well can absorb in nm?
λlongest =
1 answer:
You might be interested in
Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as
Cp-Cv= R
Cp=7/2 R
Now by putting the values
a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that
Process is adiabatic