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NEET CHEMISTRYStructure of AtomMedium

Question

In hydrogen atom, what is the de Broglie wavelength of an electron in the second Bohr orbit? [Given that Bohr radius, a0=52.9a_0 = 52.9 pm]

A

211.6 pm

B

211.6 π\pi pm

C

52.9 π\pi pm

D

105.8 pm

Step-by-Step Solution

According to the de Broglie relation and Bohr's postulate of angular momentum quantization, the circumference of the nn-th orbit is an integral multiple of the de Broglie wavelength (λ\lambda). The relationship is given by: 2πrn=nλ2\pi r_n = n\lambda where rnr_n is the radius of the nn-th orbit. The radius of the nn-th orbit in a hydrogen atom is given by the formula: rn=n2a0r_n = n^2 a_0 For the second Bohr orbit, n=2n = 2. Given a0=52.9a_0 = 52.9 pm: r2=(2)2×52.9 pm=4×52.9 pm=211.6 pmr_2 = (2)^2 \times 52.9 \text{ pm} = 4 \times 52.9 \text{ pm} = 211.6 \text{ pm} Substituting the values into the circumference equation: 2π(211.6)=2λ2\pi (211.6) = 2 \lambda λ=π×211.6 pm=211.6π pm\lambda = \pi \times 211.6 \text{ pm} = 211.6 \pi \text{ pm}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Structure of Atom. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYStructure of Atomhydrogenbrogliewavelengthelectronsecond

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