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  • Electrons with de-Broglie a fall on the target in an x-ray tube. The cut off wave length of the emitted x-rays.  Option: 1 <img alt=

Electrons with de-Broglie a fall on the target in an x-ray tube. The cut off wave length of the emitted x-rays.
 

Option: 1

\mathrm{ \lambda_0=\frac{2 m c \lambda^2}{h} }


Option: 2

\mathrm{ \lambda_0=\frac{2 h}{m c} }


Option: 3

\mathrm{\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2} }


Option: 4

\lambda_0=\lambda


Answers (1)

best_answer

Momentum of striking electrons-

\mathrm{\rho=\frac{h}{\lambda}}

\therefore kinetic energy of striking electrons.

\mathrm{k=\frac{p^2}{2 m}=\frac{\hbar^2}{2 m \lambda^2} }

This is also maximum energy of x-ray photons.

\mathrm{\frac{h c}{\partial_0}=\frac{h^2}{2 m \lambda^2} \text { or } \lambda_0=\frac{2 m \lambda^2 c}{h}}

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Riya

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