Question

6. Consider a quantum system of N particles with only three possible states to oc- cupy for each particle. The energy values of these states are equal to 0, €and €3, respectively. (a) [10 points) You observe that the probability to sample eash state is P = 0.9. P2 = 0.09 and p = 0.01 at T = 300 K. What are the energies and c? Recall that the probability to occupy state is proportional to where k= 1.38 x 10-19 J/K. (b) [10 points) What is the average energy (total) of this system?

Answer 1

6a) there are N particles and the 3 states have energy 0 ,$\varepsilon$₂,$\varepsilon$₃ and Temperature T=300K

partition function q = e⁰+ e^{−$\varepsilon$₂/kT} + e^{−$\varepsilon$₃/kT}

q=1+ e^{−$\varepsilon$₂/kT} + e^{−$\varepsilon$₃/kT}

p₁ = 0.9,P₂₌ 0.09, P3 =0.01

probability for ith level P_i = (e^{−$\varepsilon$_i/kT})/q

for i=1 or ground state energy is 0

so

P₁ = (e^−0/kT)/q

P₁ = (e⁰)/q

P₁ = 1/q

0.9=1/q

or q=1/0.9

for i=2

p₂ = (e^{−$\varepsilon$₂/kT} )*1/q

0.09 = (e^{−$\varepsilon$₂/kT} )*0.9

0.1=(e^{−$\varepsilon$₂/kT} )

taking natural logarithm on both side

ln 0.1=^{−$\varepsilon$₂/kT}

$\varepsilon$₂= kT ln 10

$\varepsilon$₂= (1,38 *10^-19 J /K )(300 K) ln 10

$\varepsilon$₂= 9.53*10^-17 J

for i=3

p₃ = (e^{−$\varepsilon$₃/kT} )*1/q

0.01 = (e^{−$\varepsilon$₃/kT} )*0.9

1/90=(e^{−$\varepsilon$₃/kT} )

taking natural logarithm on both side

ln (1/90=^{−$\varepsilon$₃/kT}

$\varepsilon$₃= kT ln (90)

$\varepsilon$₃= (1,38 *10^-19 J /K )(300 K) ln (90)

$\varepsilon$₃= 1.862*10^-16 J

b)average energy $< \varepsilon>$ is given by $\sum p_{i}\varepsilon _{i}$

$< \varepsilon>$ = $p_{1} \varepsilon_{1}&plus;p_{2} \varepsilon_{2}&plus;p_{3} \varepsilon_{3}$

$< \varepsilon>$    =$p_{1} \varepsilon_{1}&plus;p_{2} \varepsilon_{2}&plus;p_{3} \varepsilon_{3} = 0.9*0&plus;0.09*9.53*10^{-17} &plus;0.01*1.862*10^{-16} J$

$< \varepsilon>$=$8.58*10^{-18} &plus;1.862*10^{-18} J$

$< \varepsilon>$=$1.0442*10^{-17} J$