Answer:
The probability of higher energy state is 0.4200.
Explanation:
Given that,
Temperature = 247 K
Energy difference between two states [tex]E_{2}-E_{1}=1.1\times10^{-21}\ J[/tex]
We need to calculate the probability of higher energy state
Probability of [tex]E_{1}= e^{-\beta E_{1}}[/tex]
Probability of [tex]E_{2}= e^{-\beta E_{2}}[/tex]
The total probability is
[tex]e^{-\beta E_{1}}+e^{-\beta E_{1}}=1[/tex]
Here, E₁ = lower energy state
E₂ = higher energy state
Put the value of E₁ in to the formula
[tex]e^{-\beta(E_{2}-1.1\times10^{-21})}+e^{-\beta E_{1}}=1[/tex]
[tex]e^{-\beta E_{2}}(e^{\beta 1.1\times10^{-21}}+1)=1[/tex]
[tex]e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{KT}}}[/tex]
Here, [tex]\beta=\dfrac{1}{KT}[/tex]
Put the value into the formula
[tex]e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{1.380\times10^{-23}\times247}}}[/tex]
[tex]e^{\beta E_{2}}=0.4200[/tex]
Hence, The probability of higher energy state is 0.4200.
