JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 3)
What should be the order of arrangement of de-Broglie wavelength of electron ($$\lambda$$e), an $$\alpha$$-particle ($$\lambda$$a) and proton ($$\lambda$$p) given that all have the same kinetic energy?
$$\lambda$$e = $$\lambda$$p = $$\lambda$$$$\alpha$$
$$\lambda$$e < $$\lambda$$p < $$\lambda$$$$\alpha$$
$$\lambda$$e > $$\lambda$$p > $$\lambda$$$$\alpha$$
$$\lambda$$e = $$\lambda$$p > $$\lambda$$$$\alpha$$
Explanation
$$\lambda = {h \over p} = {h \over {\sqrt {2mE} }} \propto {1 \over {\sqrt m }}$$
m$$\alpha$$ > mp > me
So, $$\lambda$$e > $$\lambda$$p > $$\lambda$$$$\alpha$$
m$$\alpha$$ > mp > me
So, $$\lambda$$e > $$\lambda$$p > $$\lambda$$$$\alpha$$
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