Question

what fraction of Na atoms found in the first excited state in a sodium vapour lamp at a temperature of 300,400 and 500°c?
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Answer 1

Given that,

Temperatures are

$T_{1}=300^{\circ}C = 573 K$

$T_{2}=400^{\circ}C = 673\ K$

$T_{3}=500^{\circ}C = 773\ K$

Let N₂ be the population of first excited and N₁ be the population of the ground state

We need to calculate the fraction between the atoms in the first excited state and the ground state

Using formula of excited state

$\dfrac{N_{2}}{N_{1}}=\dfrac{e^{-\dfrac{E_{2}}{kT}}}{e^{-\dfrac{E_{1}}{kT}}}$

$\dfrac{N_{2}}{N_{1}}=e^{\dfrac{-(E_{2}-E_{1})}{kT}}$

$\dfrac{N_{2}}{N_{1}}=e^{\dfrac{-(h\nu)}{kT}}$

$\dfrac{N_{2}}{N_{1}}=e^{\dfrac{-(hc)}{\lambda kT}}$

Put the value into the formula

$\dfrac{N_{2}}{N_{1}}=exp(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.38\times10^{-23}\times573\times5.9\times10^{-7}})$

$\dfrac{N_{2}}{N_{1}}=exp(-42.63)$

$\dfrac{N_{2}}{N_{1}}=3.062\times10^{-19}$

The fraction between the atoms in the first excited state and the ground state is $3.062\times10^{-19}$

For 400°C temperature,

Using formula of the fraction

$\dfrac{N_{2}}{N_{1}}=e^{\dfrac{-(hc)}{\lambda kT}}$

$\dfrac{N_{2}}{N_{1}}=exp(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.38\times10^{-23}\times673\times5.9\times10^{-7}})$

$\dfrac{N_{2}}{N_{1}}=1.73\times10^{-16}$

For 500°C temperature,

Using formula of the fraction

$\dfrac{N_{2}}{N_{1}}=e^{\dfrac{-(hc)}{\lambda kT}}$

$\dfrac{N_{2}}{N_{1}}=exp(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.38\times10^{-23}\times773\times5.9\times10^{-7}})$

$\dfrac{N_{2}}{N_{1}}=1.88\times10^{-14}$

Hence, (I). The fraction between the atoms in the first excited state and the ground state is $3.062\times10^{-19}$

(II). The fraction between the atoms in the first excited state and the ground state is $1.73\times10^{-16}$

(III). The fraction between the atoms in the first excited state and the ground state is $1.88\times10^{-14}$