1. 10 V 2. 10² V 3. 10³ V 4. 10⁴ V

Answer: 4. 10⁴ V de Broglie wavelength of an electron accelerated through V volts: λ=1.227/sqrt V nm. Given λ=1.227×10⁻² nm: sqrt V=frac1.2271.227×10⁻²=100⇒ V=10⁴\ V Answer: option 4.

Topic: Matter waves, de Broglie relation· Dual Nature of Matter and Radiation

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10^-2 nm, the potential difference is :

1. 10 V

2. $10^2$ V

3. $10^3$ V

4. $10^4$ V

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Answer: 4. $10^4$ V

de Broglie wavelength of an electron accelerated through $V$ volts: $\lambda=\dfrac{1.227}{\sqrt V}$ nm.

Given $\lambda=1.227\times10^{-2}$ nm:

$$\sqrt V=\frac{1.227}{1.227\times10^{-2}}=100\Rightarrow V=10^4\ \text{V}$$

Answer: option 4.

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

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An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10^-2 nm, the potential difference is :

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