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Question
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
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Solution
Potential difference, V = 56 V
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
`lambda = 12.27/sqrt("V")` Å
= `12.27/sqrt56 xx 10^(-10) "m"`
= 0.1639 nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
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