Dual Nature of Matter

JEE Main level, single correct, medium

70 of 200

Question 70 of 200

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65A˚4.65\text{Å} ). The de-Broglie wavelength of this electron is:

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Answer: (C) 9.7A˚9.7\text{Å}

According to Bohr’s atomic model
mvr=nh2πmvr=\dfrac{nh}{2\pi }
And the de-Broglie wavelength
λ=hmv\lambda =\dfrac{h}{mv}
From the above two equations we can write, λ=2πrn\lambda =\dfrac{2\pi r}{n}
In the 2nd excited state
λ=2πr3\lambda =\dfrac{2\pi r}{3}
∴λ=23π×4.65Å=9.7Å∴\lambda =\dfrac{2}{3}\pi \times 4.65Å=9.7Å

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