221k views
4 votes
For a set of nondegenerate levels with energy ε/k = 0, 100 and 200 K, calculate the probability of occupying the ground level (i = 0) when T = 90 K.

P0,90K =

2) For a set of nondegenerate levels with energy ε/k = 0, 100 and 200 K , calculate the probability of occupying the excited state (i = 1) when T =90 K

P1,90K

3) For a set of nondegenerate levels with energy ε/k = 0, 100 and 200 K , calculate the probability of occupying the excited state (i = 2) when T =90 K

P2,90K=

For a set of nondegenerate levels with energy ε/k = 0, 100 and 200 K , calculate the probability of occupying the ground level (i = 0) when T =900 K

For a set of nondegenerate levels with energy ε/k = 0, 100 and 200 K , calculate the probability of occupying the excited state (i = 1) when T =900 K

1 Answer

3 votes

The probability of occupying the ground level increases as the temperature increases, while the probability of occupying the excited states decreases.

Given data: Nondegenerate energy levels with ε/k = 0, 100, and 200 K.'

The probability of occupying the ground level (i=0) when T=90 K is:

P0,90K = e^(-ε0/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }

MP0,90K = e^(-0/k × 90 K) / { e^(-0/k × 90 K) + e^(-100/k × 90 K) + e^(-200/k × 90 K) }P0,90K

= 1 / { 1 + e^(-100 × 9) + e^(-200 × 9) }= 0.9475 (approximately)

The probability of occupying the excited state (i=1)

when T=90 K is:P1,90K = e^(-ε1/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }P1,90K

= e^(-100/k × 90 K) / { e^(-0/k × 90 K) + e^(-100/k × 90 K) + e^(-200/k × 90 K) }P1,90K

= e^(-9000) / { 1 + e^(-9000) + e^(-18000) }= 0.052 (approximately)

The probability of occupying the excited state (i=2) when

T=90 K is:P2,90K = e^(-ε2/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }P2,90K

= e^(-200/k × 90 K) / { e^(-0/k × 90 K) + e^(-100/k × 90 K) + e^(-200/k × 90 K) }P2,90K

= e^(-18000) / { 1 + e^(-9000) + e^(-18000) }

= 0.0005 (approximately)

The probability of occupying the ground level (i=0) when

T=900 K is:

P0,900K = e^(-ε0/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }

P0,900K = e^(-0/k × 900 K) / { e^(-0/k × 900 K) + e^(-100/k × 900 K) + e^(-200/k × 900 K) }

P0,900K = 1 / { 1 + e^(-100 × 90) + e^(-200 × 90) }

= 0.9999999999970 (approximately)

The probability of occupying the excited state (i=1)

when T=900 K is:P1,900K = e^(-ε1/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }P1,900K

= e^(-100/k × 900 K) / { e^(-0/k × 900 K) + e^(-100/k × 900 K) + e^(-200/k × 900 K) }

P1,900K = e^(-90000) / { 1 + e^(-90000) + e^(-180000) }= 1.5 × 10^-8 (approximately)

Therefore, the probability of occupying the ground level increases as the temperature increases, while the probability of occupying the excited states decreases.

To know more about Probability, The probability of occupying the ground level increases as the temperature increases, while the probability of occupying the excited states decreases.

Given data: Nondegenerate energy levels with ε/k = 0, 100, and 200 K.'

The probability of occupying the ground level (i=0) when T=90 K is:

P0,90K = e^(-ε0/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }

MP0,90K = e^(-0/k × 90 K) / { e^(-0/k × 90 K) + e^(-100/k × 90 K) + e^(-200/k × 90 K) }P0,90K

= 1 / { 1 + e^(-100 × 9) + e^(-200 × 9) }= 0.9475 (approximately)

The probability of occupying the excited state (i=1)

when T=90 K is:P1,90K = e^(-ε1/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }P1,90K

= e^(-100/k × 90 K) / { e^(-0/k × 90 K) + e^(-100/k × 90 K) + e^(-200/k × 90 K) }P1,90K

= e^(-9000) / { 1 + e^(-9000) + e^(-18000) }= 0.052 (approximately)

The probability of occupying the excited state (i=2) when

T=90 K is:P2,90K = e^(-ε2/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }P2,90K

= e^(-200/k × 90 K) / { e^(-0/k × 90 K) + e^(-100/k × 90 K) + e^(-200/k × 90 K) }P2,90K

= e^(-18000) / { 1 + e^(-9000) + e^(-18000) }

= 0.0005 (approximately)

The probability of occupying the ground level (i=0) when

T=900 K is:

P0,900K = e^(-ε0/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }

P0,900K = e^(-0/k × 900 K) / { e^(-0/k × 900 K) + e^(-100/k × 900 K) + e^(-200/k × 900 K) }

P0,900K = 1 / { 1 + e^(-100 × 90) + e^(-200 × 90) }

= 0.9999999999970 (approximately)

The probability of occupying the excited state (i=1)

when T=900 K is:P1,900K = e^(-ε1/kT) / { e^(-ε0/kT) + e^(-ε1/kT) + e^(-ε2/kT) }P1,900K

= e^(-100/k × 900 K) / { e^(-0/k × 900 K) + e^(-100/k × 900 K) + e^(-200/k × 900 K) }

P1,900K = e^(-90000) / { 1 + e^(-90000) + e^(-180000) }= 1.5 × 10^-8 (approximately)

Therefore, the probability of occupying the ground level increases as the temperature increases, while the probability of occupying the excited states decreases.

To know more about standars d

User Andrew Guy
by
7.7k points