The Question is from neet-physics current-electricity , and it was created by the author P.NARASIMHA

Question

A metallic conductor has a current of 5A flowing through it. The free electron density is $8.5 	imes 10^{28} m^{-3}$. The cross-sectional area of the conductor is $2 	imes 10^{-6} m^2$.  If the average relaxation time is $2 	imes 10^{-14} s$ and the electric field strength inside the conductor is 2 V/m, what is the mass of an electron (approximately)?  (Assume the charge of an electron is $1.6 	imes 10^{-19} C$).

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Drift velocity and current density relation to electron properties.
	**Related Topics**: Ohm's Law, electrical conductivity, relaxation time, electron mobility.
	**Prerequisites**: Basic understanding of current, drift velocity, electric field, and relaxation time.
## SOLUTION PATHWAY
**Given Information**:
	Current (I) = 5 A
	Free electron density (n) = $8.5 	imes 10^{28} m^{-3}$
	Cross-sectional area (A) = $2 	imes 10^{-6} m^2$
	Relaxation time ($	au$) = $2 	imes 10^{-14} s$
	Electric field (E) = 2 V/m
	Charge of an electron (e) = $1.6 	imes 10^{-19} C$
**Method & Steps**:
	1.	Calculate the drift velocity (v_d) using the current equation.
		Reasoning: The current is related to the drift velocity, electron density, charge, and area.
		Formula/Rule: $I = nAe v_d$
	2.	Calculate the drift velocity.
		Working: $v_d = \frac{I}{nAe} = \frac{5}{(8.5 	imes 10^{28})(2 	imes 10^{-6})(1.6 	imes 10^{-19})} = \frac{5}{2.72 	imes 10^{4}} \approx 1.838 	imes 10^{-4} m/s$
		Key Point: Ensure consistent units.
	3.	Relate drift velocity to the electric field and relaxation time to find the mass of the electron (m).
		Reasoning: The drift velocity is also related to the electric field, relaxation time, and the mass of the electron.
		Formula/Rule: $v_d = \frac{eE	au}{m}$
	4.	Solve for the mass of the electron.
		Working: $m = \frac{eE	au}{v_d} = \frac{(1.6 	imes 10^{-19})(2)(2 	imes 10^{-14})}{1.838 	imes 10^{-4}} = \frac{6.4 	imes 10^{-33}}{1.838 	imes 10^{-4}} \approx 3.48 	imes 10^{-29} kg$
		Result: The mass of an electron is approximately $3.48 	imes 10^{-29} kg$.
		Verification: The calculated mass aligns with the correct answer.
 ## OPTION ANALYSIS
[A] :x: This is the standard mass of an electron, but the question requires calculation using given parameters.
[B] ✓: This is the correct calculated mass of the electron using the provided values.
[C] :x: This is the mass of a proton or neutron, not an electron.
[D] :x: This answer is not close to the calculated value using the given parameters and it is not a standard value.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Using the standard mass of an electron instead of calculating it using the given data. [How to avoid]: Carefully read the question and notice that it specifically asks for the mass based on the provided information.
		2.[Error 2]: Incorrect unit conversion or calculation errors when solving for drift velocity or mass. [How to avoid]: Double check the units and calculations at each step.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the relationship $v_d= \frac{eE	au}{m}$ which combines the electric field and relaxation time.
		2.[Memory aid]: Recall that current density (J) is also related to drift velocity ($J = n e v_d$) which can be useful for solving similar problems.

P.NARASIMHA · Answer

9.11 x 10^-31 kg

P.NARASIMHA · Answer

The correct Option is "3.48 x 10^-29 kg"

P.NARASIMHA · Answer

1.67 x 10^-27 kg

P.NARASIMHA · Answer

1.2 x 10^-30 kg

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