JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 5)
An $$\alpha$$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:
$${\lambda _\alpha } > {\lambda _p} < {\lambda _e}$$
$${\lambda _\alpha } > {\lambda _p} > {\lambda _e}$$
$${\lambda _\alpha } = {\lambda _p} = {\lambda _e}$$
$${\lambda _\alpha } < {\lambda _p} < {\lambda _e}$$
Explanation
$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m k}}$
So, $\lambda \propto \frac{1}{\sqrt{m}}$
So, $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$
So, $\lambda \propto \frac{1}{\sqrt{m}}$
So, $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$
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