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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
We know that the relation between `lamda` and k is given by `lamda=h/(sqrt(2mk))`
Here, for the given value of energy `k,h/(sqrt(2k))` is a constant
Thus, `lamda prop 1/sqrtm`
`:. lamda_p : lamda_n : lamda_e : lamda_alpha`
`implies 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) gt m_p` therefore `lamda_(alpha) = lamda_p`
As `m_(e) gt m_n` therefore `lamda_(e) = lamda_n`
Hence `lamda_(alpha) lt lamda_p = lamda_n lt lamda_e`
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