The wavelength of a very fast-moving electron | vedclass

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(A) According to the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the particle.For

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$\lambda = \frac{h}{m_0 v / \sqrt{1 - \frac{v^2}{c^2}}}$

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(A) According to the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the particle.For a particle moving at relativistic speeds $(v \approx c)$,the mass $m$ is given by the relativistic mass formula: $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$,where $m_0$ is the rest mass.The momentum $p$ is defined as $p = m \cdot v$.Substituting the expression for relativistic mass into the momentum formula,we get: $p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$.Finally,substituting this into the de Broglie wavelength equation,we obtain: $\lambda = \frac{h}{m_0 v / \sqrt{1 - \frac{v^2}{c^2}}}$.

The wavelength of a very fast-moving electron $(v \approx c)$ is:

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