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Byju's Answer

Standard XII

Chemistry

Thomson's Discovery of Electrons

Drift speed o...

Question

Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross-section $5 {mm}^{2}$ , is $v$ . If the electron density in copper is $9 \times 10^{28} m^{- 3}$ the value of $v$ in ${mm s}^{- 1}$ close to $(Take charge of electron to be = 1.6 \times 10^{- 19} C$ )

0.02

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Solution

The correct option is A $0.02$
Using, $I = n e A v_{d}$

$Drift speed, v_{d} = \frac{1}{n e A}$

$\Rightarrow v = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6}}$

$\Rightarrow v = 0.02 {mm s}^{- 1}$

Hence, option $(A)$ is correct.

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Similar questions

Q. Drift speed of electrons, when

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A copper wire has a cross sectional area of $1 m m^{2}$ . The electron density of copper is $8.5 \times 10^{28} m^{3}$ . Calculate the mean drift velocity of the electrons through the wire when the current is 3 A.

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Q. A current of 1.0 A exists in a copper wire of cross-section 1.0 mm². Assuming one free electron per atom, calculate the drift speed of the free electrons in the wire. The density of copper is 9000 kg m^–3.

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Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross-section 5 mm2, is v. If the electron density in copper is 9×1028 m−3 the value of v in mm s−1 close to (Take charge of electron to be=1.6×10−19 C)

The correct option is A 0.02Using, I=neAvd Drift speed, vd=1neA ⇒v=1.59×1028×1.6×10−19×5×10−6 ⇒v=0.02 mm s−1 Hence, option (A) is correct.

Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross-section $5 {mm}^{2}$ , is $v$ . If the electron density in copper is $9 \times 10^{28} m^{- 3}$ the value of $v$ in ${mm s}^{- 1}$ close to $(Take charge of electron to be = 1.6 \times 10^{- 19} C$ )

The correct option is A $0.02$
Using, $I = n e A v_{d}$

$Drift speed, v_{d} = \frac{1}{n e A}$

$\Rightarrow v = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6}}$

$\Rightarrow v = 0.02 {mm s}^{- 1}$

Hence, option $(A)$ is correct.