A proton, electron and an $\alpha$ particle are accelerated with same potential. Arrange their de-Broglies wavelengths in ascending order. A. Wavelength of proton $=\left(\lambda_p\right)$ Choose the correct answer from the options given below: |
$\lambda_\alpha<\lambda_p<\lambda_e$ $\lambda_e<\lambda_\alpha<\lambda_p$ $\lambda_e<\lambda_p<\lambda_\alpha$ $\lambda_p<\lambda_\alpha<\lambda_e$ |
$\lambda_\alpha<\lambda_p<\lambda_e$ |
The correct answer is Option (1) → $\lambda_\alpha<\lambda_p<\lambda_e$ De-Broglie wavelength of a changed. Particle moving in an electric field is given by: $i=\frac{h}{\sqrt{2 m q V}}$ $m_\alpha>m_p>m_e$ $\lambda_\alpha<\lambda_p<\lambda_e$ |