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A proton, a neutron, an electron and an ...

A proton, a neutron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare as

A

`lamda_(p)= lamda_(n) gt lamda_(e) gt lamda_(alpha)`

B

`lamda_(alpha) lt lamda_(p) = lamda_(n) gt lamda_(e)`

C

`lamda_(e) lt lamda_(p)= lamda_(n) gt lamda_(alpha)`

D

`lamda_(e )= lamda_(p)= lamda_(n)= lamda_(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
B
Matter waves (de-Broglie waves) According to de -Broglie a moving material particle sometimes acts as a wave and sometimes as a particle. De-Broglie wavelength `lamda_(d)= (h)/(p)`
`E_(p)= E_(n)= E_(e )= E_(x)`
`K.E. = K = (1)/(2)mv^(2)`
`2K= mv^(2)`
`2Km=m^(2)v^(2)`
`mK= p^(2)`
`sqrt(2mK)= p`
`therefore lamda_(d)= (h)/(p)`
`lamda_(d)= (h)/(sqrt(2mK))`
or `lamda_(d)= (h)/(sqrt(2mK))` [as h and E (K.E) is constt.]
`lamda prop (1)/(sqrtm)`
`m_(a) gt m_(p)= m_(n) gt m_(e)`
`therefore lamda_(a) lt lamda_(p)= lamda_(n) lt lamda_(e)`
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