Easy MCQ +4 / -1 PYQ · JEE Mains 2021

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?

A $\lambda$e = $\lambda$p = $\lambda$$\alpha$
B $\lambda$e < $\lambda$p < $\lambda$$\alpha$
C $\lambda$e > $\lambda$p > $\lambda$$\alpha$ Correct answer
D $\lambda$e = $\lambda$p > $\lambda$$\alpha$

Solution

$\lambda = {h \over p} = {h \over {\sqrt {2mE} }} \propto {1 \over {\sqrt m }}$ m$\alpha$ > mp > me So, $\lambda$e > $\lambda$p > $\lambda$$\alpha$

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Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect

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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?

Solution

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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?