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NEET PHYSICSMedium

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (me=9×10−31m_e = 9 \times 10^{-31} kg)

1

12.2×10−1312.2 \times 10^{-13} m

2

12.2×10−1212.2 \times 10^{-12} m

3

12.2×10−1412.2 \times 10^{-14} m

4

12.2 nm

Step-by-Step Solution

For an electron accelerated through a potential V, λ=12.27V A˚=12.27×10−1010000=12.27×10−12\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA} = \frac{12.27 \times 10^{-10}}{\sqrt{10000}} = 12.27 \times 10^{-12} m

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