Hall VoltageHardQuestion 20209
Question
A semiconductor strip of thickness 2 mm carries current 5 A in a field B = 0.3 T (perpendicular to strip). Hall voltage measured is 15 mV. Carrier density n is (q = 1.6×10⁻¹⁹):
- A3.1×10²² m⁻³Correct
- B3.1×10²⁴ m⁻³
- C6.25×10²² m⁻³
- D1.56×10²² m⁻³
Correct answer
3.1×10²² m⁻³
Explanation
VH = IB/(ned) → n = IB/(VH×e×d) = (5×0.3)/(0.015×1.6×10⁻¹⁹×0.002) = 1.5/(4.8×10⁻²⁴) ≈ 3.1×10²³ m⁻³. (Check: this is a typical semiconductor value.)