A proton, electron and an $\alpha$ particle are accelerated with same potential. Arrange their de-Broglies wavelengths in ascending order.
A. Wavelength of proton $=\left(\lambda_p\right)$
B. Wavelength of electron $=\left(\lambda_e\right)$
C. Wavelength of particle $=\left(\lambda_\alpha\right)$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
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:
$i=\frac{h}{\sqrt{2 m q V}}$
$m_\alpha>m_p>m_e$
$\lambda_\alpha<\lambda_p<\lambda_e$