Correct option is B
Solution:
σ = - 1, 0, 1
H=−μBhσ+Δ(1−σ2). Draw energy levels using σ=−1,0,1 and H=−μBhσ+Δ(1−σ2) Z=eβμBh+e−βμBh+e−βΔF=−kBTln(eβμBh+e−βμBh+e−βΔ)dF=−SdT−PdV−μdh μ=(∂h∂F)T,V=kBTeβμBh+e−βμBh+e−βΔeβμBh−e−βμBh⋅βμB =kBTβμBeβμBh+e−βμBh+e−βΔeβμBh−e−βμBh In the limit where h is small M=2+e−βΔ(1+βμBh−1+βμBh)μB =2+e−βΔβμB2