A proton, a neutron, an electron and an α -particle have same energy. Then, their de-Broglie wavelengths compare as:
Answer & explanation
Correct answer: option 2
We know that the relation between λ and K is given by $λ=\frac{h}{\sqrt{2mK}}$
Here, for the given value of energy K, $\frac{h}{\sqrt{2K}}$ is a constant.
Thus, $λ∝\frac{1}{\sqrt{m}}$
$∴λ_p:λ_n:λ_e:λ_α$
$⇒=\frac{1}{\sqrt{m_p}}:\frac{1}{\sqrt{m_n}}:\frac{1}{\sqrt{m_e}}:\frac{1}{\sqrt{m_α}}$
Since, $m_p=m_n$, hence $λ_p=λ_n$
As, $m_α<m_p$, therefore $λ_α>λ_n$
Hence, $λ_α<λ_p=λ_n<λ_e$