1. 2.25×10¹⁵ 2. 2.5×10⁶ 3. 2.5×10⁻⁶ 4. 2.25×10⁻¹⁵

Answer: 2. 2.5×10⁶ Mobility is drift velocity per unit electric field: μ=v_d/E . μ=frac7.5×10⁻⁴3×10⁻¹⁰=2.5×10⁶\ m²V⁻¹s⁻¹ Answer: option 2.

A charged particle having drift velocity of 7.5×10^-4 m s^-1 in an electric field of 3×10^-10 V m^-1, has a mobility in m^2 V^-1 s^-1 of :

1. $2.25\times10^{15}$

2. $2.5\times10^{6}$

3. $2.5\times10^{-6}$

4. $2.25\times10^{-15}$

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

Best answer

Answer: 2. $2.5\times10^{6}$

Mobility is drift velocity per unit electric field: $\mu=\dfrac{v_d}{E}$.

$$\mu=\frac{7.5\times10^{-4}}{3\times10^{-10}}=2.5\times10^{6}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$$

Answer: option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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