91Ó°ÊÓ

We have been given that the emission spectrum of sodium has a wavelength of $499nm$in $5d→3p$transition. We need to find energy of $5d$state.

We are given with $\mathrm{λ}=499nm$and $E_{3p}=2.104eV$, where $\mathrm{λ}$is wavelength and $E_{3p}$is energy of $3p$state. We also know about energy equation , which can be written as :

localid="1650300462496" $E_{5d→3p}=\frac{hc}{\mathrm{λ}}$,localid="1650300525081" $\mathrm{λ}$here is wavelength and we know the standard value of localid="1650300529926" $h$and localid="1650300534619" $c$

By substituting the values we get,

$E_{5d→3p}=\frac{1240eVnm}{499nm}=2.485eV$

To find the value of localid="1650300538851" $E_{5d}$,we can use the equation ,localid="1650300543949" $E_{5d}=E_{3p}+E_{5d→3p}$

We will substitute the values on right hand side of the equation,

localid="1650300548403" $E_{5d}=2.104eV+2.485eV=4.589eV$