Question

Find the value of four quantum numbers for the 24th electron of iron.​

Answer 1

Answer:

then n=3 , l=2 (because it is the d orbital), ml is −2 (because you count up one the series from −l to +l six times, for the six electrons in the d subshell) and ms .

I hope it May helps to you

Answer 2

To Find :-

• Value of $\sf \bigg(a{}^{2}+\dfrac{1}{a{}^{2}}\bigg) =?$

Given :-

• $\sf a=3+\sqrt{8}$

$\sf Then$

$\sf\bigg(\dfrac{1}{a}\bigg)=\dfrac{1}{3+\sqrt{8}}$

$\sf:\implies \dfrac{1}{3+\sqrt{8}}\times \dfrac{(3-\sqrt{8})}{(3-\sqrt{8})}$

$\sf:\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{(3){}^{2}-(\sqrt{8}){}^{2}}$

$\sf:\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{9-8}$

$\sf:\implies \dfrac{1}{a}= 3-\sqrt{8}$

$\sf Now$

$\sf \bigg(a{}^{2}+\dfrac{1}{a{}^{2}}\bigg)$

$\sf:\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}+2$

$\sf:\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=( 3+\cancel{\sqrt{8}}+3-\cancel{\sqrt{8}}){}^{2}$

$\sf:\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}= (6){}^{2}=9$

$\sf:\implies 9=a{}^{2}+\dfrac{1}{a{}^{2}}+2$

$\sf:\implies 9-2=a{}^{2}+\dfrac{1}{a{}^{2}}$

$\sf:\implies 7=a{}^{2}+\dfrac{1}{a{}^{2}}$

$\sf:\implies a{}^{2}+\dfrac{1}{a{}^{2}}= 7$