An electron is accelerated through a potential difference $V$. According to the relation given in the notes, its de Broglie wavelength is
A$\lambda = \dfrac{12.29}{V}$
B$\lambda = \dfrac{12.29}{\sqrt{V}}$
C$\lambda = 12.29\sqrt{V}$
D$\lambda = \dfrac{h}{eV}$
Answer & Solution
Correct answer: B. $\lambda = \dfrac{12.29}{\sqrt{V}}$
For an electron accelerated through potential difference $V$, the gained kinetic energy is $eV=\tfrac12 mv^2$. Using $\lambda=h/(mv)$ gives $\lambda=\dfrac{h}{\sqrt{2meV}}$. In the form quoted in the notes, this becomes $\lambda = 12.29/\sqrt{V}$ when suitable units are used.
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