Electrons of mass m with de-Broglie wavelengt | vedclass

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(A) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.Since the kinetic energy $E

Vedclass · Accepted Answer

$\frac{2 m c \lambda^2}{h}$

Vedclass · Answer

(A) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.Since the kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the de-Broglie equation: $\lambda = \frac{h}{\sqrt{2mE}}$.Squaring both sides: $\lambda^2 = \frac{h^2}{2mE}$,which gives $E = \frac{h^2}{2m\lambda^2}$.The cut-off wavelength $\lambda_0$ of the emitted $X$-rays corresponds to the maximum energy of the photon,which is equal to the kinetic energy $E$ of the incident electron: $E = \frac{hc}{\lambda_0}$.Equating the two expressions for $E$: $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.Solving for $\lambda_0$: $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on a target. The cut-off wavelength $\lambda_0$ is equal to [$h=$ Planck's constant,$c=$ velocity of light].

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