CUET Preparation Today
A proton and $\alpha$-particle have same kinetic energy. The de-Broglie wavelength ratio $\frac{\lambda_{p}}{\lambda_\alpha}$ is equal to: |
$\frac{m_\alpha}{m_{p}}$ $m_\alpha \times m_{p}$ $\sqrt{\frac{m_\alpha}{m_p}}$ $\sqrt{\frac{m_{p}}{m_\alpha}}$ |
$\sqrt{\frac{m_\alpha}{m_p}}$ |
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] $⇒λ=\frac{h}{\sqrt{2mK}}$ $[p=\sqrt{2mK}]$ $⇒λ_p=\frac{h}{\sqrt{2m_pK}},λ_α=\frac{h}{\sqrt{2m_αK}}$ $\frac{λ_p}{λ_α}=\sqrt{\frac{m_α}{m_p}}$ |
