Measurement of Properties: Tensile Test, Hardness, Electrical Transport and Thermal Analysis

This chapter answers the remaining paragraphs of Section XE2.4 of XE2 Materials Science, one of the optional sections of the Engineering Sciences (XE) paper: “Tensile test: engineering and true stress-strain curves, parameters, such as, yield stress, ultimate tensile stress, elongation, area under the curve; Hardness: Brinell, Rockwell and Vickers. Electrical conductivity, carrier mobility and concentrations. Thermal analysis techniques: thermogravimetry and calorimetry.” It teaches how the engineering and true curves are built and converted, the parameters read from them, resilience and toughness as areas, the three hardness tests, the four-point probe and the Hall measurement, and what thermogravimetry and calorimetry curves show. The numericals are true stress and strain, necking strain, resilience, Brinell number, four-point-probe resistivity, Hall mobility and carrier density, and thermogravimetric mass balance.

The tensile test and the engineering stress-strain curve

A standard specimen of original cross-section A₀ and gauge length L₀ is pulled at a slow constant rate while load F and extension ΔL are recorded. The engineering stress is σ = F/A₀ and the engineering strain is e = ΔL/L₀; both use the original dimensions. The initial straight line has slope E (Young’s modulus). Curvature begins at the proportional or elastic limit; for metals without a sharp yield point the yield stress is defined by the 0.2 % offset method: draw a line of slope E starting from e = 0.002 and read the stress where it cuts the curve.

Parameters read from the engineering curve
ParameterDefinitionRemark
Yield stress σ_yStress at 0.2 % offset plastic strainOnset of significant plastic flow
Ultimate tensile stress (UTS)F_max/A₀, the maximum of the curveNecking starts here in a ductile metal
Percent elongation(L_f − L₀)/L₀ × 100, measured on the broken piecesElastic part has recovered; depends on gauge length
Percent reduction in area(A₀ − A_f)/A₀ × 100Measure of ductility not tied to gauge length
⚠️ Trap: yield point, UTS and fracture stress
The UTS is the highest point of the engineering curve, not the stress at which the specimen breaks. In a ductile metal the engineering stress at fracture is lower than the UTS because the load falls during necking while A₀ is still used in the division. Unloading from any plastic point follows a line parallel to the elastic slope, so the permanent strain is the intercept of that line on the strain axis.

True stress and true strain, and necking

The true stress divides the load by the instantaneous area, σ_t = F/A, and the true strain adds up the increments, ε_t = ∫dL/L = ln(L/L₀). Plastic flow conserves volume, A L = A₀ L₀, as long as the deformation is uniform along the gauge. Then A = A₀/(1 + e) and σ_t = σ(1 + e), ε_t = ln(1 + e). These conversions are valid only up to the UTS; after necking starts the strain is concentrated in the neck and the gauge-length ratio no longer describes it (the neck area must be measured instead, ε_t = ln(A₀/A)).

🧠 Worked example
At e = 0.15 the engineering stress is 400 MPa. σ_t = 400 × 1.15 = 460 MPa and ε_t = ln 1.15 = 0.140. Since ln(1 + e) < e, true strain is always smaller than engineering strain for tension, while true stress is larger. Check: 0.140 < 0.15 and 460 > 400.

In the plastic range many metals follow σ_t = K ε_t^n, where n is the strain-hardening exponent. The load is F = σ_t A. Necking starts when an increase in strain hardening no longer compensates the loss of area, dF = 0, i.e. dσ_t/dε_t = σ_t (the Considère criterion). For the power law this gives ε_t = n at necking, equivalently e = eⁿ − 1 in engineering strain. After the UTS the engineering stress falls because the load falls, yet the true stress still rises: the fall is geometric, not a loss of strength.

⚠️ Trap: strains and additivity
True strains add: stretching 100 mm to 110 mm and then to 121 mm gives ln 1.1 + ln 1.1 = ln 1.21. Engineering strains do not: 10 % then 10 % is 21 %, not 20 %. For compressive deformation the sign of ε_t is negative and ln(1 + e) with e < 0 is still correct.

Area under the curve: resilience and toughness

The area under the stress-strain curve is energy absorbed per unit volume (Pa × 1 = J/m³). Up to yield, with σ = Ee, it is a triangle: the modulus of resilience U_r = ½ σ_y e_y = σ_y²/2E, the energy recovered on unloading. The area up to fracture is the toughness (tensile toughness), the energy needed to break a unit volume. For a ductile metal it is roughly the average of σ_y and UTS times the strain to fracture; for a brittle material the curve ends at a small strain, so the area is small.

🧠 Worked example
A steel has σ_y = 300 MPa and E = 200 GPa. U_r = (300 × 10⁶)²/(2 × 200 × 10⁹) = 9 × 10¹⁶/(4 × 10¹¹) = 2.25 × 10⁵ J/m³ = 225 kJ/m³. Check: σ_y²/2E has units Pa²/Pa = Pa = J/m³. Doubling σ_y quadruples U_r; doubling E halves it, which is why a high-yield, low-modulus material (a spring alloy) stores the most elastic energy.
⚠️ Trap: strength is not toughness
A hard, strong but brittle material can have a smaller area under the curve than a soft, ductile one. Compare the areas, not the heights. Also the area under the engineering curve and under the true curve differ; the toughness of the question is the engineering one unless true quantities are stated.

Hardness: Brinell, Rockwell and Vickers

Hardness is resistance to permanent indentation. Brinell presses a hard ball of diameter D with load P (commonly D = 10 mm and P = 3000 kgf for steels) and measures the diameter d of the circular impression: HB = 2P/[πD(D − √(D² − d²))] in kgf/mm², the load divided by the curved area of the cap. A valid test keeps d between 0.25D and 0.5D. Rockwell is depth-based: a minor load (10 kgf) seats the indenter, a major load is applied and removed, and the permanent extra depth h is converted to a number, HR = N − h/0.002 mm, with N = 100 for diamond-cone scales and 130 for ball scales; a deeper dent means a softer sample and a lower number. Vickers uses a square-based diamond pyramid with 136° between opposite faces and measures the mean diagonal d of the impression: HV = 2F sin 68°/d² = 1.854 F/d² (F in kgf, d in mm).

The three hardness tests
TestIndenter and loadWhat is measuredNotes
Brinell HBBall 10 mm, 500 to 3000 kgfImpression diameterLarge impression, averages coarse structure; not for very hard steel
Rockwell C (HRC)Diamond cone 120°, 10 kgf minor, 150 kgf majorDepth increaseFast, direct reading; hardened steel; Rockwell B uses a 1/16 in ball and 100 kgf
Vickers HVDiamond pyramid 136°, loads from grams (micro-hardness) to about 100 kgfMean diagonalSelf-similar impressions, so HV is nearly independent of load; one scale for soft to hard
🧠 Worked example and correlations
A 10 mm ball under 3000 kgf leaves an impression of d = 3.0 mm. √(100 − 9) = 9.539, D − 9.539 = 0.461 mm, πD × 0.461 = 14.47 mm², HB = 6000/14.47 = 415 kgf/mm². For many steels the tensile strength in MPa is about 3.45 × HB, and for a ductile metal the indentation pressure is roughly three times the flow stress; these are rough rules, not exact laws, and they hold only when the stem scopes them.

Electrical conductivity and the four-point probe

Conductivity is σ = 1/ρ, where ρ = RA/L for a uniform bar. In a semiconductor σ = e(nμ_e + pμ_h): carrier densities n and p and mobilities μ = v_d/E, the drift speed per unit field (m²/V s). A two-probe measurement adds the contact and lead resistances to the sample’s, which can dominate for a semiconductor or a low-resistance metal. The four-point probe removes them: the outer two probes pass a known current I, and the inner two, connected to a high-impedance voltmeter that draws almost no current, read the voltage V; no current flows through the voltage contacts, so their contact drops do not enter.

For four collinear probes with equal spacing s: on a sample much larger than s in all directions (semi-infinite), ρ = 2π s V/I. On a thin sheet of thickness t much smaller than s and lateral extent much larger than s, ρ = (π/ln 2) t V/I = 4.532 t V/I, and the sheet resistance is R_s = ρ/t = 4.532 V/I (ohm per square). Finite-size and edge corrections are applied for real samples.

🧠 Worked example
On a silicon wafer 0.25 mm thick the probe reads V = 5.0 mV at I = 2.0 mA (thin-sheet case). ρ = 4.532 × (0.25 × 10⁻³ m) × (5.0 × 10⁻³/2.0 × 10⁻³) = 4.532 × 2.5 × 10⁻⁴ × 2.5 = 2.83 × 10⁻³ Ω m = 0.283 Ω cm. Check units: m × (V/A) = m × Ω = Ω m. The thin-sheet and bulk formulas are different; using 2πsV/I on a thin film overstates ρ because the current is confined to the film.

Hall measurement: carrier type, concentration and mobility

Pass a current I along a slab and apply a magnetic field B perpendicular to the current. The Lorentz force pushes carriers sideways until a transverse electric field balances it, producing the Hall voltage across the width. With a single carrier type of density n and charge q, and t the slab thickness measured along B, V_H = I B/(n q t). The Hall coefficient R_H = V_H t/(I B) = 1/(n q): it is negative for electrons and positive for holes, so its sign gives the carrier type and its magnitude gives the density, n = 1/(e|R_H|).

Combine with a resistivity measurement (for example from the four-point probe) to find the Hall mobility μ_H = |R_H| σ = |R_H|/ρ, because σ = n e μ. For mixed conduction R_H = (pμ_h² − nμ_e²)/[e(pμ_h + nμ_e)²]; in an intrinsic semiconductor n = p, so the sign follows whichever mobility is larger (usually electrons).

🧠 Worked example
A slab of thickness t = 0.20 mm carries I = 1.0 mA in a field B = 0.50 T and shows |V_H| = 2.5 mV. Use e = 1.602 × 10⁻¹⁹ C. n = IB/(e t V_H) = (1.0 × 10⁻³ × 0.50)/(1.602 × 10⁻¹⁹ × 2.0 × 10⁻⁴ × 2.5 × 10⁻³) = 5.0 × 10⁻⁴/8.01 × 10⁻²⁶ = 6.24 × 10²¹ m⁻³. A metal has n near 10²⁸ m⁻³, so its V_H would be millions of times smaller, which is why Hall tests are done on semiconductors or thin films.
⚠️ Trap: which dimension is t
In V_H = IB/(nqt) the thickness t is the dimension parallel to B, not the width across which V_H is measured. Using the width cancels incorrectly and gives a wrong n. Also, the sign is a single-carrier result; a small R_H of either sign in a mixed sample does not mean few carriers.

Thermal analysis: thermogravimetry and calorimetry

Thermogravimetric analysis (TGA) records the mass of a sample against temperature (or time) in a controlled atmosphere at a fixed heating rate. A step down is a mass loss (dehydration, decomposition, evaporation); a rise is a mass gain (oxidation in air). The derivative curve (DTG) peaks at the fastest-loss temperature. A transition with no mass change, such as melting of a pure metal in an inert gas, is invisible to TGA. The step height is converted to composition by stoichiometry.

Calorimetry here means differential scanning calorimetry (DSC): the heat flow needed to keep sample and reference on the same temperature programme is recorded against temperature. The peak area is proportional to the enthalpy change ΔH of the transition: endothermic peaks (melting, evaporation, dissociation) and exothermic peaks (crystallisation, curing, oxidation) point in opposite directions, and which way is plotted up depends on the instrument, so the axis label decides. A glass transition shows as a step in the baseline because the heat capacity changes, with no latent heat and no peak. The heat capacity follows from the baseline offset, C_p = (heat flow)/(m × heating rate). Differential thermal analysis (DTA) is the sibling method that records the temperature difference between sample and reference.

🧠 Worked example
Calcium carbonate decomposes in an inert gas as CaCO₃ → CaO + CO₂. With atomic masses Ca = 40, C = 12, O = 16, the molar masses are 100 and 56 g/mol for CaCO₃ and CaO. A 10.0 mg sample leaves a residue of 10.0 × 56/100 = 5.6 mg, a loss of 44 %. Check: the loss equals the CO₂ fraction, 44/100.

Key takeaways

  • Engineering σ = F/A₀, e = ΔL/L₀; yield by 0.2 % offset; UTS is F_max/A₀; elongation and reduction in area measure ductility.
  • True: σ_t = σ(1+e), ε_t = ln(1+e), valid to the UTS only; necking starts at ε_t = n for σ_t = Kε_t^n; true strains add, engineering strains do not.
  • Area under the curve is energy per unit volume: resilience σ_y²/2E up to yield, toughness up to fracture; strong does not mean tough.
  • HB = 2P/[πD(D − √(D²−d²))]; Rockwell reads depth (HR = N − h/0.002 mm); HV = 1.854F/d² with F in kgf and d in mm.
  • Four-point probe removes contact resistance: ρ = 2πsV/I (bulk) and ρ = 4.532 tV/I (thin sheet).
  • Hall: n = IB/(e t V_H), the sign of R_H gives carrier type, μ_H = |R_H|/ρ; TGA tracks mass, DSC tracks heat flow and shows Tg as a baseline step.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In a tensile test of a ductile metal, the ultimate tensile stress (UTS) is

    1. the engineering stress at which the specimen finally breaks apart
    2. the maximum load divided by the original cross-section
    3. the stress at which the first plastic strain appears
    4. the true stress recorded at the instant of final fracture
    Show answer

    Answer: B — the maximum load divided by the original cross-section

    UTS is F_max/A₀, the highest point of the engineering curve, reached at the onset of necking. Fracture occurs later at a lower engineering stress because the load falls during necking. The first plastic strain marks the elastic or yield limit, and the true stress at fracture is a different, larger quantity.

  2. A metal shows no sharp yield point. Its yield stress is then conventionally found by

    1. taking the engineering stress recorded at the instant the specimen fractures, as the yield point
    2. taking the stress at the highest point of the engineering stress-strain curve
    3. drawing a horizontal line at a stress equal to 0.2 % of the UTS and reading its intersection
    4. drawing a line of the elastic slope from 0.002 strain and reading where it cuts the curve
    Show answer

    Answer: D — drawing a line of the elastic slope from 0.002 strain and reading where it cuts the curve

    The 0.2 % offset method draws a line of slope E from e = 0.002; its intersection with the curve is the offset yield stress, which is the stress that leaves 0.2 % permanent strain on unloading. Fracture and UTS are different points, and the 0.2 % applies to strain, not to a fraction of the UTS.

  3. Beyond the UTS the engineering stress of a ductile metal falls, yet the true stress keeps rising. The correct explanation is that

    1. the load falls as the neck shrinks faster than hardening raises stress, and engineering stress uses the original area
    2. the metal softens after the UTS through internal damage, so its true flow stress really decreases
    3. the true stress is computed with the original area and the load, and the load keeps rising with strain
    4. volume is no longer conserved in the neck, so the cross-sectional area actually increases while the load drops
    Show answer

    Answer: A — the load falls as the neck shrinks faster than hardening raises stress, and engineering stress uses the original area

    In the neck the true stress F/A still rises because strain hardening continues, but the instability criterion dF = 0 has passed, so the load F falls as A shrinks faster. Engineering stress F/A₀ follows the load and therefore falls. Softening is not the cause, the true stress uses the instantaneous (not original) area, and plastic flow conserves volume.

  4. A ductile metal obeys σ_t = K ε_t^n in the plastic range with n = 0.25. Using the condition for the onset of necking, the engineering strain at the UTS is closest to

    1. 0.250
    2. 0.125
    3. 0.284
    4. 0.500
    Show answer

    Answer: C — 0.284

    Necking starts at true strain ε_t = n = 0.25. Converting, e = exp(ε_t) − 1 = e^0.25 − 1 = 0.284. The value 0.250 is the true strain mistaken for the engineering strain; because ln(1 + e) < e, the engineering strain is larger than the true strain.

  5. Which description of the Rockwell C hardness test is correct?

    1. A 10 mm hardened steel ball pressed under 3000 kgf, with the diameter of the impression measured afterwards
    2. A diamond cone under 10 kgf minor and 150 kgf major loads, depth converted to a number
    3. A square-based diamond pyramid whose mean diagonal is measured under a load of a few grams
    4. A hardened ball under 100 kgf with the number equal to the load divided by the cap area
    Show answer

    Answer: B — A diamond cone under 10 kgf minor and 150 kgf major loads, depth converted to a number

    Rockwell C uses a diamond cone of 120° cone angle with a 10 kgf minor load and a 150 kgf major load, and the number comes from the extra depth, HR = 100 − h/0.002 mm. The ball with an impression diameter belongs to Brinell, the pyramid with a diagonal to Vickers, and load over cap area is the Brinell definition.

  6. In a Vickers test, HV = 1.854 F/d² with F in kgf and d in mm. A load of 30 kgf leaves an impression of mean diagonal 0.40 mm. The Vickers hardness is closest to

    1. 139
    2. 696
    3. 348
    4. 87
    Show answer

    Answer: C — 348

    HV = 1.854 × 30/(0.40)² = 55.62/0.16 = 347.6, about 348. The value 139 comes from dividing by d instead of d²; 87 comes from using d² = 0.64; 696 from doubling the result.

  7. A semiconductor bar has contact resistances of several ohms at its ends, much larger than the bar’s own resistance. Why does a four-point probe still give the correct resistivity?

    1. It drives a much higher current through the bar, which makes the contact resistance negligible compared with the sample
    2. It measures the voltage at the current-carrying outer contacts and then subtracts the known contact drop
    3. It uses alternating current, for which a metal-semiconductor contact has no resistance at all in practice
    4. The voltage is read across two inner contacts by a high-impedance meter, so they carry almost no current
    Show answer

    Answer: D — The voltage is read across two inner contacts by a high-impedance meter, so they carry almost no current

    The current enters and leaves through the outer probes, so their contact resistance only affects how much voltage the source must supply, not the measured V. The inner probes feed a voltmeter of very high input impedance, so the current through them, and the drop across their contacts, is negligible. A larger current does not remove contact resistance, and the method does not subtract anything.

  8. For an intrinsic semiconductor (n = p) with μ_e = 3μ_h, use R_H = (pμ_h² − nμ_e²)/[e(pμ_h + nμ_e)²]. The Hall coefficient is

    1. negative, since the more mobile electrons dominate
    2. zero, because equal numbers of electrons and holes cancel exactly
    3. positive, because holes carry positive charge and matter most
    4. undefined, because the denominator vanishes when n equals p
    Show answer

    Answer: A — negative, since the more mobile electrons dominate

    With n = p the numerator is n(μ_h² − μ_e²) = n μ_h²(1 − 9) < 0, so R_H is negative; the denominator e n²(μ_h + μ_e)² is positive and finite. R_H would vanish only if μ_e = μ_h. Equal densities do not cancel the Hall effect because the weighting is by mobility squared, not by number.

  9. A DSC heating scan of an amorphous polymer shows a step-like shift of the baseline, with no peak. This feature is best attributed to

    1. melting of small crystallites, which absorbs latent heat and gives a broad dip
    2. a glass transition, where heat capacity changes without latent heat
    3. a mass loss from thermal decomposition of the polymer chain
    4. crystallisation, which releases latent heat as an exothermic peak
    Show answer

    Answer: B — a glass transition, where heat capacity changes without latent heat

    A glass transition is a change in heat capacity with no latent heat, so DSC shows a baseline step rather than a peak. Melting and crystallisation give endothermic and exothermic peaks with areas proportional to ΔH. A mass loss is seen by TGA, not DSC.

  10. Consider uniform plastic deformation of a metal rod in tension, before necking. Which statements are correct?

    1. P: The true strain is ln(1 + e), where e is the engineering strain.
    2. Q: The relation σ_t = σ(1 + e) remains valid right up to fracture, including the neck.
    3. R: Two successive engineering strains of 10 % each give a total engineering strain of 20 %.
    4. S: Two successive true strains of 0.10 each give a total true strain of 0.20.
    Show answer

    Answer: A — P: The true strain is ln(1 + e), where e is the engineering strain.; D — S: Two successive true strains of 0.10 each give a total true strain of 0.20.

    P holds from dε_t = dL/L. S holds because true strains add (ln is additive). Q fails: the relation uses A = A₀/(1 + e), which requires uniform deformation and breaks once a neck forms. R fails: 1.1 × 1.1 = 1.21, so the total engineering strain is 21 %.

  11. Which statements about hardness testing are correct?

    1. P: Brinell hardness is obtained from the diameter of the impression left by a hard ball.
    2. Q: Rockwell hardness is read from the permanent increase in indentation depth produced by the major load (measured under the minor load), without optical measurement.
    3. R: A higher HRC number means a softer material because the dent is deeper.
    4. S: In the Vickers formula HV = 1.854 F/d², d is the depth of the pyramid indentation.
    Show answer

    Answer: A — P: Brinell hardness is obtained from the diameter of the impression left by a hard ball.; B — Q: Rockwell hardness is read from the permanent increase in indentation depth produced by the major load (measured under the minor load), without optical measurement.

    P and Q describe Brinell and Rockwell correctly. R is reversed: HR = N − h/0.002 mm, so a deeper dent gives a lower number and a higher HRC means a harder material. In S, d is the mean diagonal of the impression, not its depth.

  12. Which statements about thermal analysis are correct?

    1. P: Thermogravimetry records the change of sample mass with temperature or time.
    2. Q: The area of a DSC peak is proportional to the enthalpy change of the transition.
    3. R: DSC detects a mass loss directly because it weighs the pan.
    4. S: Melting of a pure metal in an inert gas appears as a step in a TGA curve.
    Show answer

    Answer: A — P: Thermogravimetry records the change of sample mass with temperature or time.; B — Q: The area of a DSC peak is proportional to the enthalpy change of the transition.

    P defines TGA and Q states the calorimetric relation. R is false: DSC records heat flow, not mass. S is false: melting in an inert atmosphere changes no mass (and vapour loss is negligible well below the boiling point), so a TGA curve stays flat.

  13. In a tensile test the engineering stress is 320 MPa at an engineering strain of 0.25, before necking begins. Assuming uniform deformation and constant volume, find the true stress in MPa. Give the answer as an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 400

    σ_t = σ(1 + e) = 320 × 1.25 = 400 MPa. Check: A = A₀/1.25, so the same load acts on a smaller area and the true stress is 25 % larger.

  14. A steel has yield stress 250 MPa and Young’s modulus 200 GPa. Find its modulus of resilience in kJ/m³, taking the elastic part of the curve as linear. Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 156.25

    U_r = σ_y²/2E = (250e6)²/(2 × 200e9) = 6.25e16/4e11 = 1.5625e5 J/m³ = 156.25 kJ/m³. Cross-check: ½ × 250e6 × (250e6/200e9) = ½ × 250e6 × 1.25e-3 = 1.5625e5 J/m³.

  15. A Brinell test uses a 10 mm ball with a load of 3000 kgf and leaves an impression of diameter 4.0 mm. Use HB = 2P/[πD(D − √(D² − d²))] in kgf/mm². Find the Brinell hardness number, as an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 229

    √(D² − d²) = √(100 − 16) = 9.165 mm; D − 9.165 = 0.835 mm; πD × 0.835 = 26.23 mm²; HB = 6000/26.23 = 228.8, i.e. 229. Python check: 2·3000/(π·10·(10 − √84)) = 228.77.

  16. A thin silicon slice of thickness 0.50 mm, very large in lateral extent compared with the probe spacing, is measured with a collinear four-point probe. The current is 4.0 mA and the voltage between the inner probes is 8.0 mV. Use ρ = (π/ln 2) t V/I for a thin sheet. Find the resistivity in Ω cm, to three decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.453

    V/I = 8.0 mV/4.0 mA = 2.0 Ω. π/ln 2 = 4.532. ρ = 4.532 × 0.50 × 10⁻³ m × 2.0 Ω = 4.532 × 10⁻³ Ω m = 0.4532 Ω cm (1 Ω m = 100 Ω cm). Python check gives 0.45324.

  17. An n-type slab of thickness t = 0.10 mm (measured along the field) carries I = 2.0 mA in a magnetic field B = 0.40 T. The Hall voltage is 4.0 mV and the resistivity is 2.0 × 10⁻³ Ω m. Take R_H = V_H t/(IB) and μ_H = |R_H|/ρ for a single carrier type. Find the Hall mobility in m²/(V s), to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.25

    R_H = V_H t/(IB) = (4.0e-3 × 1.0e-4)/(2.0e-3 × 0.40) = 4.0e-7/8.0e-4 = 5.0e-4 m³/C. μ_H = 5.0e-4/2.0e-3 = 0.25 m²/(V s), i.e. 2500 cm²/(V s). Python check: 0.25.