Question
de Broglie wavelength associated with an electron accelerated through a potential difference V $\lambda$ is . What will be its wavelength when the accelerating potential is increased to 4 V?
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Answer
$\lambda = \frac{\text{h}}{\sqrt{2\text{meV}}}$ or $\frac{12.27}{\sqrt{\text{V}}}\text{A}^{\circ}$
$\lambda' = \frac{12.27}{\sqrt{4\text{V}}}\text{A}^{\circ}$ or $\frac{\lambda}{2}$.
$\lambda' = \frac{12.27}{\sqrt{4\text{V}}}\text{A}^{\circ}$ or $\frac{\lambda}{2}$.
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