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Current Electricity question

2019 · 9 Apr · Shift 2 · Q45
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Current Electricity question

2019 · 9 Apr · Shift 2 · Q45

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is 25ƒs (femto second), it's approximate resistivity is :- (me = 9.1 × 10–31 kg)
  1. A
    10–8 Ω\OmegaΩ m
  2. B
    10–7 Ω\OmegaΩ m
  3. C
    10–5 Ω\OmegaΩ m
  4. D
    10–6 Ω\OmegaΩ m
View written solutionFree

Correct answer: A

  1. Use Drude model relation

For a conductor,

σ=ne2τme\sigma = \frac{n e^2 \tau}{m_e}σ=me​ne2τ​

so resistivity is

ρ=1σ=mene2τ.\rho = \frac{1}{\sigma} = \frac{m_e}{n e^2 \tau}.ρ=σ1​=ne2τme​​.
  1. Given data
  • Number density of electrons: n=8.5×1028 m−3n = 8.5 \times 10^{28}\ \text{m}^{-3}n=8.5×1028 m−3
  • Mean free time: τ=25 fs=25×10−15 s\tau = 25\ \text{fs} = 25 \times 10^{-15}\ \text{s}τ=25 fs=25×10−15 s
  • Electron mass: me=9.1×10−31 kgm_e = 9.1 \times 10^{-31}\ \text{kg}me​=9.1×10−31 kg
  • Electron charge: e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}e=1.6×10−19 C
  1. Substitute into the formula
ρ=9.1×10−31(8.5×1028)(1.6×10−19)2(25×10−15)\rho = \frac{9.1 \times 10^{-31}}{(8.5 \times 10^{28})(1.6 \times 10^{-19})^2(25 \times 10^{-15})}ρ=(8.5×1028)(1.6×10−19)2(25×10−15)9.1×10−31​

First,

(1.6×10−19)2=2.56×10−38(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}(1.6×10−19)2=2.56×10−38

Now the denominator:

8.5×2.56×25×1028−38−158.5 \times 2.56 \times 25 \times 10^{28-38-15}8.5×2.56×25×1028−38−15 =544×10−25= 544 \times 10^{-25}=544×10−25 =5.44×10−23= 5.44 \times 10^{-23}=5.44×10−23

Thus,

ρ=9.1×10−315.44×10−23\rho = \frac{9.1 \times 10^{-31}}{5.44 \times 10^{-23}}ρ=5.44×10−239.1×10−31​ =9.15.44×10−8= \frac{9.1}{5.44} \times 10^{-8}=5.449.1​×10−8 ≈1.67×10−8 Ω m\approx 1.67 \times 10^{-8}\ \Omega\,\text{m}≈1.67×10−8 Ωm
  1. Approximate value
ρ≈10−8 Ω m\rho \approx 10^{-8}\ \Omega\,\text{m}ρ≈10−8 Ωm
  1. Check options
  • A: 10−8 Ω10^{-8}\ \Omega10−8 Ω m ✅
  • B: 10−7 Ω10^{-7}\ \Omega10−7 Ω m ❌
  • C: 10−5 Ω10^{-5}\ \Omega10−5 Ω m ❌
  • D: 10−6 Ω10^{-6}\ \Omega10−6 Ω m ❌

Hence, the correct option is A.

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