Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 A). The de-Broglie wavelength of this electron is: (a) 6.6 A (b) 3.5 A (c) 9.7 A (d) 12.9 A

Answer: (c). Second excited state: n = 3. λ = (2π r)/(n) = (2×3.14×4.65)/(3) A λ ≈ 9.7 A So the answer is (c) 9.7 A.

Consider an electron in a hydrogen atom revolving in its second… | JEE Main 2019 (12 Apr, Shift 2)

Q102JEE Main 2019 · 12 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum

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

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Answer: (c)

Second excited state: $n = 3$ .
$\lambda = \frac{2\pi r}{n} = \frac{2\times3.14\times4.65}{3}\ \mathring{A}$
$\lambda \approx 9.7\ \mathring{A}$
So the answer is (c) $9.7\ \mathring{A}$ .

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