Answer to Question #185637 in Molecular Physics | Thermodynamics for Neha Nirmal

Question

A system has N distinguishable particles. Each particle has two non-degenerate states 

with level separation of 0.15 eV. Calculate the average number of the particles in each 

state when the system is in thermal equilibrium with the bath temperature of 300 K.

(3)

Answer 1

At  thermal equilibrium ,number of particles in each state is given by,

$n_1 = {N\over Z} e^{−ε_1/k_BT }= {N\over 1+ e^{ −ε/k_BT}}$

$n_2= {N\over Z} e^{−ε_2/k_BT }= {N\over 1+ e^{ −ε/k_BT}}$

where, ε₁=energy of first state and ε₂=energy of second state

and

ε =energy difference between the two states(level separation)

=0.15 eV

T=300 k

k_B= Boltzmann constant

now,

ratio of no. of particles in the two states,

${n_1 \over n_2}=e^{{(ε_2 -ε_1)}\over k_B T}$

${n_1 \over n_2}=e^{{(ε)}\over k_B T}$

${n_1 \over n_2}=e^{{((0.15)1.60×10^{−19})}\over 1.38 × 10^{−23} (300)}$

${n_1 \over n_2}=e^{5.79}$

${n_1 \over n_2}=327.01$