An electron is accelerated through a potential difference of 10,000 V . Its de Broglie wavelength is, (nearly): ( m_e = 9 × 10^-31 kg ) - Physics MCQ NEET 2019 2026 | ExamTest.live

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An electron is accelerated through a potential difference of 10, 000 V . Its de Broglie wavelength is, (nearly): (m_{e} = 9 \times 1 0^{- 31} kg)

Solution & Step-by-step Explanation

For an electron accelerated through potential V, the de Broglie wavelength (λ) is given by: λ = \frac{12.27}{V} \overset{˚}{A} Given V = 10, 000 V, so V = 100 . λ = \frac{12.27}{100} \overset{˚}{A} = 0.1227 \overset{˚}{A} = 12.27 \times 1 0^{- 12} m This is approximately 12.2 \times 1 0^{- 12} m .

Try it yourself before checking the explanation above.

An electron is accelerated through a potential difference of 10, 000 V . Its de Broglie wavelength is, (nearly): (m_{e} = 9 \times 1 0^{- 31} kg)

12.2 \times 1 0^{- 13} m

12.2 \times 1 0^{- 12} m

12.2 \times 1 0^{- 14} m

12.2 nm

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