ChemistryIn hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Give that Bohr radius, a0=52.9pm ]

In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is $[Give that Bohr radius, a_{0} = 52.9 pm]$

A
105.8 pm
B
211.6 pm
C
$211.6 π pm$
D
$52.9 π pm$

Solution:

$2 π r_{n} = n λ$

$For n = 2, r = r_{0} \times \frac{(2)^{2}}{(1)}$

$So, 2 \times π \times 52.9 \times 4 = 2 \times λ$

$λ = 211.6 π pm$

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