The de-Broglie wavelength of an electron in the first Bohr orbit is

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equal to one-fourth the circumference of the first orbit

equal to half the circumference of first orbit

equal to twice the circumference of first orbit

equal to the circumference of the first orbit

answer is D.

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Detailed Solution

According to Bohr the circumference of the orbital will be integral multiple of de-Broglie wavelength.

$\Rightarrow nλ = 2 {πr}_{n}$

or first orbit n=1 so λ=2πr₁.

$λ$ equals to the circumference of the first orbit.

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