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Structure of Atom question

2023 · 6 Apr · Shift 2 · Q5

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Structure of Atom
2023 · 6 Apr · Shift 2 · Q5

Structure of Atom question

2023 · 6 Apr · Shift 2 · Q5

JEE MainChemistryStructure of AtomMCQ+4 / −1

If the radius of the first orbit of hydrogen atom is

\alpha_{0}

, then de Broglie's wavelength of electron in

3^{\text {rd }}

orbit is :

A
$\frac{\pi \alpha^{3}}{6}$
B
$3\pi\alpha_0$
C
$6\pi\alpha_0$
D
$\frac{\pi \alpha^{3}}{3}$

View written solutionFree

Correct answer: C

Use Bohr’s quantization condition

For an electron in the $n^{\text{th}}$ orbit of hydrogen atom,

$2\pi r_n = n\lambda$

where:

$r_n$ = radius of the $n^{\text{th}}$ orbit
$\lambda$ = de Broglie wavelength of the electron in that orbit

So,

$\lambda = \frac{2\pi r_n}{n}$

Write the radius of the $n^{\text{th}}$ orbit

For hydrogen atom,

$r_n = n^2\alpha_0$

Given that the radius of the first orbit is $\alpha_0$ , therefore for the $3^{\text{rd}}$ orbit,

$r_3 = 3^2\alpha_0 = 9\alpha_0$

Find the de Broglie wavelength in the 3rd orbit

Using

$\lambda = \frac{2\pi r_3}{3}$

Substitute $r_3 = 9\alpha_0$ :

$\lambda = \frac{2\pi(9\alpha_0)}{3}$

$\lambda = 6\pi\alpha_0$

Match with the options

The correct option is:

$\boxed{6\pi\alpha_0}$

So, Option C is correct.

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