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V3Ga critical temperature and fields

Short answer

Using H_c(T)=H_0[1−(T/T_c)^2] and the data at 14 K (0.176 T) and 13 K (0.528 T), solving the ratio yields T_c=14.47 K, H_c(0)=2.73 T, and H_c(4.2 K)=2.50 T.

For a specimen of V₃Ga, the critical fields are 0.176T and 0.528T for 14K and 13K respectively. Calculate the critical temperature. Also calculate critical fields at 0K and 4.2K.

The critical field varies with temperature as

Hc(T)=H0[1−T2Tc2]H_c(T) = H_0\left[1 - \frac{T^2}{T_c^2}\right]

Given: Hc1=0.176H_{c1} = 0.176 T at T1=14T_1 = 14 K, Hc2=0.528H_{c2} = 0.528 T at T2=13T_2 = 13 K.

Critical temperature

Divide the two equations:

0.1760.528=Tc2−142Tc2−13213=Tc2−196Tc2−169Tc2−169=3Tc2−5882Tc2=419Tc2=209.5Tc=14.47 K\begin{aligned} \frac{0.176}{0.528} &= \frac{T_c^2 - 14^2}{T_c^2 - 13^2} \\ \frac{1}{3} &= \frac{T_c^2 - 196}{T_c^2 - 169} \\ T_c^2 - 169 &= 3T_c^2 - 588 \\ 2T_c^2 &= 419 \\ T_c^2 &= 209.5 \\ T_c &= 14.47\ \text{K} \end{aligned}

Critical field at 0 K

H0=Hc(14)1−196209.5=0.176×209.513.5=2.731 T\begin{aligned} H_0 &= \frac{H_c(14)}{1 - \frac{196}{209.5}} = \frac{0.176 \times 209.5}{13.5} \\ &= 2.731\ \text{T} \end{aligned}

Check with the 13 K value: 2.731×(1−169/209.5)=2.731×0.1933=0.5282.731 \times (1 - 169/209.5) = 2.731 \times 0.1933 = 0.528 T, which matches.

Critical field at 4.2 K

Hc(4.2)=2.731[1−4.22209.5]=2.731 (1−0.0842)=2.501 T\begin{aligned} H_c(4.2) &= 2.731\left[1 - \frac{4.2^2}{209.5}\right] \\ &= 2.731\,(1 - 0.0842) \\ &= 2.501\ \text{T} \end{aligned}

Answer: Tc=14.47T_c = 14.47 K, Hc(0)=2.73H_c(0) = 2.73 T, Hc(4.2 K)=2.50H_c(4.2\ \text{K}) = 2.50 T.

✓ Answer checked by Claude Opus 5.5 (high effort)

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