The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

2 \pi \times 0.529 \mathop \rm A\limits^ \circ

0.529 \mathop \rm A\limits^ \circ

0.529 \over 2\pi \mathop \rm A\limits^ \circ

4 \times 0.529 \mathop \rm A\limits^ \circ

Structure of Atom PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q21 with Solution

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Correct answer: B

Use Bohr’s quantization condition

For an electron in the Bohr orbit, $2\pi r_n = n\lambda$ where $r_n$ is the radius of the $n$ th orbit and $\lambda$ is the de-Broglie wavelength.
For the first Bohr orbit

In hydrogen atom, for the first orbit: $n=1, \quad r_1 = 0.529\,\text{Å}$

So, $2\pi r_1 = \lambda$
Substitute the value of $r_1$

$\lambda = 2\pi \times 0.529\,\text{Å}$
Match with the options

This corresponds to: Option B: $2\pi \times 0.529\,\text{Å}$
Check other options briefly
- A: $0.529\,\text{Å}$ is just the radius, not the wavelength.
- C: $\dfrac{0.529}{2\pi}\,\text{Å}$ is incorrect.
- D: $4\times 0.529\,\text{Å}$ is also incorrect.

Therefore, the correct answer is B.