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  • Tell me? - Dual Nature of Matter and Radiation - NEET

Electrons of mass m with de-Broglie wavelength \lambda fall on the target in an X-ray tube. The cutoff wavelength (\lambda0) of the emitted X-ray is

  • Option 1)

    \lambda _0 = \frac{{2mc\lambda ^2 }} {h}

  • Option 2)

    \lambda _0 = \frac{{2h}} {{mc}}

  • Option 3)

    \lambda _0 = \frac{{2m^2 c^2 \lambda ^3 }} {{h^2 }}

  • Option 4)

    \lambda _0 = \lambda

 

Answers (1)

As we learnt in

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 

 

 maximum energy associated with electron = \frac{p^{2}}{2m}

p=\frac{h}{\lambda } \ \ \Rightarrow E= \frac{h^{2}}{2m\lambda ^{2}}....(1)

Let cut off wavelength is \lambda o Then energy of X ray = \frac{hc}{\lambda o}.....(2)

\Rightarrow \frac{h}{2m \lambda ^{2}}=\frac{c}{\lambda o}.....(2)

\Rightarrow \lambda o=\frac{2mc \lambda ^{2}}{h}

 


Option 1)

\lambda _0 = \frac{{2mc\lambda ^2 }} {h}

This option is correct.

Option 2)

\lambda _0 = \frac{{2h}} {{mc}}

This option is incorrect.

Option 3)

\lambda _0 = \frac{{2m^2 c^2 \lambda ^3 }} {{h^2 }}

This option is incorrect.

Option 4)

\lambda _0 = \lambda

This option is incorrect.

Posted by

Sabhrant Ambastha

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