A protom and an alpha particle are both accelerated through a potential difference of 100 V. the ratio of the de-Broglie wavelength associated with the progom to that associated with the alpha particle is

A protom and an alpha particle are both accelerated through a potentia

1.	A protom and an alpha particle are both accelerated through a potential difference of 100 V. the ratio of the de-Broglie wavelength associated with the progom to that associated with the alpha particle is
Answer» `1:1` `sqrt2:1` `2:1` `2sqrt2:1` SOLUTION :`lamda_("de-Broglie")=(H)/(sqrt(2mqV))`, HENCE for same accelerating potential potential V, we have `(lamda_(p))/(lamda_(alpha))=sqrt((m_(prop)q_(prop))/(m_(p)q_(p)))=sqrt(4xx2)=2sqrt(2)implies lamda_(p):lamda_(alpha)=2sqrt(2):1`.

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A protom and an alpha particle are both accelerated through a potential difference of 100 V. the ratio of the de-Broglie wavelength associated with the progom to that associated with the alpha particle is

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