Question

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electrons of mass M with de Broglie wavelength Lambda fall on the target an x-ray tube the cutoff wavelength of the emitted x-ray is

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Answer 1

Answer: The cutoff wavelength of the emitted x-ray is $\lambda_{0}=\dfrac{2mc\lambda^2}{h}$.

Explanation:

Given that,

Mass of electron = M

Wavelength = $\lambda$

We know that,

The momentum is

$P=\sqrt{2mE_{k}}$

Where,

$E_{k}$ = The kinetic energy  of the incident electron

The wave length is

$\lamda = \dfrac{h}{P}$.....(I)

Put the value of P in equation (I)

$\lambda = \dfrac{h}{\sqrt{2mE_{k}}}$

$E_{k}= \dfrac{h^2}{2m\lambda^2}$

We know that,

$\dfrac{hc}{\lambda_{0}}= E_{k}$

Where, h = plank's constant

c = speed of light

The cutoff wavelength of the emitted x-ray is

$\lambda_{0}=\dfrac{2mc\lambda^2}{h}$

Hence, The cutoff wavelength of the emitted x-ray is $\lambda_{0}=\dfrac{2mc\lambda^2}{h}$.