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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)gtlamda_(e)gtlamda_(alpha)`

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B
(b) : Kinetic energy of particle, `K=(1)/(2)mv^(2)`
or `mv=sqrt(2mK)`
de Bloglie wavelength, `lamda=(h)/(mv)=(h)/(sqrt2mK)`
`"For the given value of K",lamdaprop(1)/(sqrt(2mK))`
`:.lamda_(p):lamda_(n):lamda_(e):lamda_(alpha)=(1)/(sqrt(m_(p))):(1)/(sqrt(m_(n))):(1)/(sqrt(m_(e))):(1)/(sqrt(m_(alpha)))`
`"Since "m_(p)=m_(n),"hence "lamda_(p)=lamda_(n)`
`"As "m_(alpha)gtm_(p),"therefore"lamda_(alpha)ltlamda_(p)`
`"As " m_(e)ltm_(n),"therefore"lamda_(e)gtlamda_(n)`
`"Hence "lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)`
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