Question #285541

For hydrogen, an electron is in 2p state. Obtain the magnitude of orbital angular momentum and possible



z-components of orbital angular momentum 𝐿⃗ . What would be the possible values of total angular



momentums (𝐽 ) and associated magnetic moments 𝜇𝐽



⃗⃗⃗

1
Expert's answer
2022-01-12T08:32:08-0500

For P

State

l=1,s=12\frac{1}{2}

Calculate J value

L-s to L+s

Total anguler momentum

J=12,32J=\frac{1}{2},\frac{3}{2}

Z- components of anguler momentum

ml= 32,12,12,32\frac{-3}{2},\frac{-1}{2},\frac{1}{2},\frac{3}{2}

J=L12\\J=L-\frac{1}{2}

We know that formula total angular moment




μ=glp(j12)+gsp2\mu=g_{lp}(j-\frac{1}{2})+\frac{g_{sp}}{2}

glp=1,gsp=5.585μNg_{lp}=1,g_{sp}=5.585\mu_N

μ=1×(3212)+5.5852\mu=1\times(\frac{3}{2}-\frac{1}{2})+\frac{5.585}{2}

total angular



momentum (𝐽 ) and associated magnetic moment


μ=1+2.7925=3.7925μN\mu=1+2.7925=3.7925\mu_N

J=L12J=L-\frac{1}{2}


μ=jj+1(glp(j+32)gsp2)\mu=\frac{j}{j+1}(g_{lp}(j+\frac{3}{2})-\frac{g_{sp}}{2})

μ=1212+1(1×(12+32)5.5852)\mu=\frac{\frac{1}{2}}{\frac{1}{2}+1}(1\times(\frac{1}{2}+\frac{3}{2})-\frac{5.585}{2})



μ=13(25.5852)=0.264μN\mu=\frac{1}{3}(2-\frac{5.585}{2})=-0.264\mu_N


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