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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
For a particle of mass m and velocity v , kinetic energy is given by
`K = 1/2 mv^2 = p^2/(2m) " " [p=mv]`
`implies ` The de Broglie wavelength associated with the particle is
`lamda=h/p=h/(sqrt(2mK))`
`implies lamda prop 1/sqrtm`
`:. lamda_p:lamda_n: lamda_e:lamda_alpha = 1/sqrt(m_p): 1/sqrt(m_n):1/sqrt(m_e):1/sqrt(m_alpha)`
We know that `m_p = m_alpha`
`:. lamda_p=lamda_n`
Also, `m_alpha gt m_p`
`:. lamda_(alpha) lt lamda_p`
`:. lamda_(alpha) lt lamda_p = lamda_n lt lamda_e`
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