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NCERT Exemplar · Q20

Q.What are the dimensions of χ\chi, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ\chi, upto a constant by constructing a quantity of dimensions of χ\chi, out of parameters of the atom: ee, mm, vv, RR and μ0\mu_0. Here, mm is the electronic mass, vv is electronic velocity, RR is Bohr radius. Estimate the number so obtained and compare with the value of χ∼10−5\chi \sim 10^{-5} for many solid materials.

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χ\chi is dimensionless. The only dimensionless combination of {e,m,v,R,μ0}\{e,m,v,R,\mu_0\} is χ∼μ0e2/(mR)\chi \sim \mu_0 e^2/(mR) (the velocity vv drops out); numerically ≈6.7×10−4∼10−4\approx 6.7\times10^{-4} \sim 10^{-4}, about an order of magnitude larger than the observed ∼10−5\sim10^{-5}.

Dimensions of χ\chi

Susceptibility is defined by M⃗=χ H⃗\vec{M} = \chi\,\vec{H}. Magnetisation and magnetising field are both in A/m\text{A/m}, so their ratio is a pure number:

[χ]=[M][H]=A⋅m−1A⋅m−1=1(dimensionless).[\chi] = \frac{[M]}{[H]} = \frac{\text{A·m}^{-1}}{\text{A·m}^{-1}} = 1 \quad(\text{dimensionless}).

Constructing χ\chi from atomic parameters

Look for χ∼μ0aebmcvdRf\chi \sim \mu_0^{a} e^{b} m^{c} v^{d} R^{f}. In terms of M,L,T,IM,L,T,I:

[μ0]=MLT−2I−2,  [e]=IT,  [m]=M,  [v]=LT−1,  [R]=L.[\mu_0]=MLT^{-2}I^{-2},\ \ [e]=IT,\ \ [m]=M,\ \ [v]=LT^{-1},\ \ [R]=L.

Requiring the product to be dimensionless gives

M:a+c=0I:−2a+b=0T:−2a+b−d=0L:a+d+f=0\begin{aligned} M:&\quad a + c = 0\\ I:&\quad -2a + b = 0\\ T:&\quad -2a + b - d = 0\\ L:&\quad a + d + f = 0 \end{aligned}

From the II-equation b=2ab = 2a; substituting into the TT-equation gives d=0d = 0; then c=−ac = -a and f=−af = -a. The combination is therefore (μ0e2mR)a\left(\dfrac{\mu_0 e^2}{m R}\right)^{a}, and the fundamental dimensionless quantity is

χ∼μ0e2mR.\chi \sim \frac{\mu_0 e^2}{m R}.

Tip

The velocity power came out d=0d=0: vv drops out entirely. The estimate depends only on the orbit size RR and the charge-to-mass ratio, not on how fast the electron moves.

Dimension check: (MLT−2I−2)(I2T2)(M)(L)=1.\dfrac{(MLT^{-2}I^{-2})(I^2T^2)}{(M)(L)} = 1.

Numerical estimate

For the hydrogen atom (Bohr radius R=a0R = a_0):

μ0e2mR=(4π×10−7)(1.6×10−19)2(9.1×10−31)(5.3×10−11).\frac{\mu_0 e^2}{mR} = \frac{(4\pi\times10^{-7})(1.6\times10^{-19})^2}{(9.1\times10^{-31})(5.3\times10^{-11})}.

  • Numerator: 4π×10−7×2.56×10−38≈3.22×10−444\pi\times10^{-7} \times 2.56\times10^{-38} \approx 3.22\times10^{-44}.
  • Denominator: 9.1×10−31×5.3×10−11≈4.82×10−419.1\times10^{-31} \times 5.3\times10^{-11} \approx 4.82\times10^{-41}. …

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