Dual Nature of Matter

JEE Main level, single correct, medium

27 of 200

Question 27 of 200

The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume mp=me×1849{m}_{p}={m}_{e}\times 1849)

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Answer: (A) 1:431:43

The de-Broglie wavelength is given by
λ=hp=h2mK(As,p2=2mK)\lambda =\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}}(\text{As},{p}^{2}=2mK).
It is given that
(K)e=(K)p(K{)}_{e}=(K{)}_{p}
Thus, the ratio of the wavelengths of the proton and electron is

λpλe=2Kme2Kmp=memp=11849=143\dfrac{{\lambda }_{p}}{{\lambda }_{e}}=\dfrac{\sqrt{2K{m}_{e}}}{\sqrt{2K{m}_{p}}}=\sqrt{\dfrac{{m}_{e}}{{m}_{p}}}=\sqrt{\dfrac{1}{1849}}=\dfrac{1}{43}.

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