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  • The potential difference applied to an X-ray tube is V. The ratio of de Broglie wavelength of an electron to the minimum wavelength of the X-ray is directly proportional to <div class='qna

The potential difference applied to an X-ray tube is V. The ratio of de Broglie wavelength of an electron to the minimum wavelength of the X-ray is directly proportional to 

Option: 1

V

 

 

 


Option: 2

\sqrt{V}


Option: 3

V^{3/2}


Option: 4

V^{1/2}


Answers (1)

best_answer

As we learned

 

De - Broglie wavelength with charged particle -

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

\lambda = \frac{h}{\sqrt{2mqv}}
 

- wherein

E\rightarrow kinetic\: energy\: o\! f particle

q\rightarrow charged \: particle

 

 

\lambda_e = \frac{h}{\sqrt{2meV}}\Rightarrow \lambda _e \ \alpha \ \frac{1}{\sqrt{V}} \\ for \ X-ray \\ \frac{hc}{\lambda_{min}}=eV\Rightarrow \lambda _{min } \ \alpha \ \frac{1}{V}

 So   \frac{\lambda_e }{\lambda _{min}} \ \alpha \ \frac{V}{\sqrt{V}}\\ \Rightarrow \frac{\lambda_e }{\lambda _{min}} \ \alpha \ \sqrt{V}

Posted by

Sanket Gandhi

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