A proton and $\alpha$-particle have same kinetic energy. The de-Broglie wavelength ratio $\frac{\lambda_{p}}{\lambda_\alpha}$ is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\sqrt{\frac{m_\alpha}{m_p}}$
De-Broglie Wavelength (λ) is -
$λ=\frac{h}{p}$ [p = Momentum of Particle]
$⇒λ=\frac{h}{\sqrt{2mK}}$ $[p=\sqrt{2mK}]$
$⇒λ_p=\frac{h}{\sqrt{2m_pK}},λ_α=\frac{h}{\sqrt{2m_αK}}$
$\frac{λ_p}{λ_α}=\sqrt{\frac{m_α}{m_p}}$