An α -particle and proton have same kinetic energy, the ratio of de Broglie wavelength of α -particle and proton is:
Answer & explanation
Correct answer: option 3
Kinetic energy K is related with momentum (p) of the particle by the relation, $K=\frac{p^2}{2m}$
or $p=\sqrt{2mK}$
∴ de Broglie wavelength $λ=\frac{h}{p}=\frac{h}{\sqrt{2mK}}$;
So $λ∝\frac{1}{\sqrt{m}}$
$∴\frac{λ_α}{λ_p}=\sqrt{\frac{m_P}{m_α}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$