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MCQEasyJEE 2023Dual Nature of Matter & de Broglie Relation

JEE Chemistry 2023 Question with Solution

If the radius of the first orbit of hydrogen atom is a0a_0, then de Broglie’s wavelength of electron in 33rd orbit is:

  • A

    πa06\dfrac{\pi a_0}{6}

  • B

    πa03\dfrac{\pi a_0}{3}

  • C

    6πa06\pi a_0

  • D

    3πa03\pi a_0

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The radius of the first orbit of hydrogen atom is a0a_0.

Find: The de Broglie wavelength of electron in the 33rd orbit.

Use the de Broglie relation mentioned in the solution hint:

2πr=nλ2\pi r = n\lambda

For the nnth orbit of hydrogen atom, the radius is:

rn=n2a0r_n = n^2 a_0

So for the 33rd orbit,

r3=32a0=9a0r_3 = 3^2 a_0 = 9a_0

Now apply

2πr3=3λ2\pi r_3 = 3\lambda

Therefore,

λ=2πr33=2π⋅9a03=6πa0\lambda = \frac{2\pi r_3}{3} = \frac{2\pi \cdot 9a_0}{3} = 6\pi a_0

Therefore, the de Broglie wavelength is 6πa06\pi a_0. The correct option is C.

Orbit Formula Shortcut

Given: The first orbit radius is a0a_0.

Find: The wavelength in the 33rd orbit.

Using rn=n2a0r_n = n^2 a_0 and 2πrn=nλ2\pi r_n = n\lambda,

λ=2πn2a0n=2πna0\lambda = \frac{2\pi n^2 a_0}{n} = 2\pi n a_0

For n=3n = 3,

λ=2π⋅3a0=6πa0\lambda = 2\pi \cdot 3 a_0 = 6\pi a_0

This shortcut works because substituting the Bohr radius relation directly into the de Broglie condition reduces the expression to a simple formula in nn. Hence, the correct option is C.

Common mistakes

  • Using r3=3a0r_3 = 3a_0 instead of r3=9a0r_3 = 9a_0. This is wrong because Bohr orbit radius varies as n2n^2, not nn. Always use rn=n2a0r_n = n^2 a_0 before substituting.

  • Using 2πr=λ2\pi r = \lambda for every orbit. This is incorrect because the standing-wave condition is 2πr=nλ2\pi r = n\lambda. The orbit number nn must be included.

  • Matching the numerical result with the wrong option by overlooking the factor of 22 or 33. This happens when algebra is rushed. First compute λ=2πrnn\lambda = \frac{2\pi r_n}{n} carefully, then compare with the options.

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