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NEET PhysicsMedium

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 nm1.227 \times 10^{-2} \text{ nm}, the potential difference is :

1

10 \text{ V}

2

10^2 \text{ V}

3

10^3 \text{ V}

4

10^4 \text{ V}

Step-by-Step Solution

λ=12.27V A˚\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}. V=12.27×10−101.227×10−11=102\sqrt{V} = \frac{12.27 \times 10^{-10}}{1.227 \times 10^{-11}} = 10^2. Therefore, V=104 voltsV = 10^4 \text{ volts}.

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