Question

Suppose a two-level system is in a bath with temperature 247 K. The energy difference between the two states is 1.1 × 10-21 J. What is the probability that we will find the system in the higher energy state?

Answer 1

Answer:

The probability of higher energy state is 0.4200.

Explanation:

Given that,

Temperature = 247 K

Energy difference between two states $E_{2}-E_{1}=1.1\times10^{-21}\ J$

We need to calculate the probability of higher energy state

Probability of $E_{1}= e^{-\beta E_{1}}$

Probability of $E_{2}= e^{-\beta E_{2}}$

The total probability is

$e^{-\beta E_{1}}+e^{-\beta E_{1}}=1$

Here, E₁ = lower energy state

E₂ = higher energy state

Put the value of E₁ in to the formula

$e^{-\beta(E_{2}-1.1\times10^{-21})}+e^{-\beta E_{1}}=1$

$e^{-\beta E_{2}}(e^{\beta 1.1\times10^{-21}}+1)=1$

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{KT}}}$

Here, $\beta=\dfrac{1}{KT}$

Put the value into the formula

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{1.380\times10^{-23}\times247}}}$

$e^{\beta E_{2}}=0.4200$

Hence, The probability of higher energy state is 0.4200.