An electron of mass m when accelerated t...

An electron of mass `m` when accelerated through a potential difference `V` has de - Broglie wavelength `lambda`. The de - Broglie wavelength associated with a proton of mass `M` accelerated through the same potential difference will be

`(lamda)/(3)sqrt((M)/(m))`

`(lamda)/(3).(M)/(m)`

`(lamda)/(3)sqrt((m)/(M))`

`(lamda)/(3).(m)/(M)`

Text Solution

Verified by Experts

The correct Answer is:

`lamda=(h)/(sqrt(2mqV))`
`:'` q is constant.
`:.lamdaprop(1)/(sqrt(mV))`
`(lamda_(p))/(lamda_(e))=sqrt((m_(e)V_(e))/(m_(p)V_(p)))=sqrt((mV)/(M9V))`
`(lamda_(p))/(lamda_(e))=sqrt((m)/(M))xx(1)/(3)`
`lamda_(p)=(lamda_(e))/(3)sqrt((m)/(M))=(lamda)/(3)sqrt((m)/(M))`.

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An electron of mass m, when accelerated through a potential difference V, has de-Broglie wavelength lamda , the de-Broglie wavelength associated with a protom of mass M and accelerated through the same potential difference V will be

`lamda(m)/(M)`

`lamdasqrt((m)/(M))`

`lamdasqrt((M)/(m))`

`(lamdaM)/(m)`

An electron of mass 'm', when accelerated through a potential V has de-Broglie wavelength lambda . The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:

`lambdasqrt(M/m)`

`lambdasqrt(m/M)`

`lambda(M/m)`

`lambda(m/M)`

The ratio of de-Broglie wavelength associated with a proton and a mu -particle accelerated through the same potential difference 'V' is

`1:2`

`1:1`

`1 : 2√2`

`1:4`