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An alpha- particle and a photon are accelerated from rest through the same potential difference V. Find the ratio of de Broglie wavelength associated with them.

Answer» <html><body><p></p>Solution :`K.E. = ( 1)/( 2 ) m upsilon^(2) = <a href="https://interviewquestions.tuteehub.com/tag/qv-611767" style="font-weight:bold;" target="_blank" title="Click to know more about QV">QV</a>` or <br/> `upsilon= sqrt((2qV)/(m))` <br/> de Broglie <a href="https://interviewquestions.tuteehub.com/tag/wavelength-1450414" style="font-weight:bold;" target="_blank" title="Click to know more about WAVELENGTH">WAVELENGTH</a>,` <a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a> = ( h )/( mv)` <br/>For same <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> difference, <br/>`( lambda_(<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>))/( lambda_(p))= sqrt((m_(p)q_(p))/( m_(alpha)q_(alpha)))=sqrt((m_(p)e)/( 4m_(p) 2e)) = (1)/( 2sqrt( 2))`</body></html>


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