A proton and an alpha-particle are accel...

A proton and an `alpha`-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths `lambda_(p) and lambda_(a)` related to each other?

Text Solution

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For a particle of mass m , charge q , accelerated through a potential V, the de Broglie wavelength is given by
`lamda=h/(sqrt(2mqV))`
For fixed V ,
`lamdaprop1/(sqrt(mq))`
Ratio of the Broglie wavelength for proton and alpha particle is
`lamda_p/lamda_(alpha)=(sqrt(m_alphaq_alpha))/(sqrt(m_pq_p))`
We know that `m_alpha=4m+p,q_p=e and q_(alpha)=2e`
`:.lamda_p/lamda_alpha=sqrt(4m_pxx2e)/sqrt(m_pxxe)=sqrt8/1`

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de-Broglie wavelength lambda is

proportional to mass

proportional to impulse

inversely proportional to impulse

independent to impulse

A proton and an electron are accelerated by same potential difference have de-Broglie wavelength lambda_(p) and lambda_(e ) .

`lambda_(e ) = lambda_(p)`

`lambda_( c ) lt lambda_(p )`

`lambda_( e ) gt lambda_( p ) `

none of these

An electron and a proton are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths will be

`(m_(p)/m_(e))^(1//2)`

`m_(e)//m_(p)`

`m_(p)//m_(e)`