Logo
Jeeify
MCQEasyJEE 2025Dual Nature of Matter & de Broglie Relation

JEE Chemistry 2025 Question with Solution

If a0a_0 is denoted as the Bohr radius of the hydrogen atom, then what is the de-Broglie wavelength λ\lambda of the electron present in the second orbit of hydrogen atom? (nn: any integer)

  • A

    4na0πa0\frac{4n a_0}{\pi a_0}

  • B

    8πa0n\frac{8 \pi a_0}{n}

  • C

    4πa0n\frac{4 \pi a_0}{n}

  • D

    2a0nπ\frac{2 a_0}{n \pi}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: a0a_0 is the Bohr radius and the electron is in the second orbit of hydrogen atom.

Find: The de-Broglie wavelength λ\lambda.

Use the standing-wave condition for an electron in Bohr orbit:

2πr=nλ2\pi r = n\lambda

So,

λ=2πrn\lambda = \frac{2\pi r}{n}

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

rn=n2a0r_n = n^2 a_0

Substituting this into the wavelength expression:

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

For the second orbit, n=2n = 2:

λ=2π(2)a0=4πa0\lambda = 2\pi (2)a_0 = 4\pi a_0

This agrees with the listed option written as:

λ=4πa0n\lambda = \frac{4\pi a_0}{n}

Therefore, the correct option is C.

Using de-Broglie and Bohr quantization

Given:

  • de-Broglie relation:
λ=hmv\lambda = \frac{h}{mv}
  • Bohr quantization condition:
mvr=nh2πmvr = \frac{nh}{2\pi}

Find: The de-Broglie wavelength for the electron in the second orbit.

From Bohr quantization,

v=nh2πmrv = \frac{nh}{2\pi mr}

Substitute into de-Broglie relation:

λ=hm(nh2πmr)=2πrn\lambda = \frac{h}{m\left(\frac{nh}{2\pi mr}\right)} = \frac{2\pi r}{n}

Now use the radius of the nnth Bohr orbit:

r=n2a0r = n^2 a_0

Hence,

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

For the second orbit, n=2n = 2:

λ=4πa0\lambda = 4\pi a_0

the solution concludes with option C, which is written as 4πa0n\frac{4\pi a_0}{n}. Therefore, the correct option is C.

Common mistakes

  • Using r=a0r = a_0 for the second orbit is incorrect because a0a_0 is the radius of the first Bohr orbit. For the nnth orbit, use rn=n2a0r_n = n^2 a_0.

  • Confusing the standing-wave condition as 2πr=λ2\pi r = \lambda is wrong. A stable orbit contains nn wavelengths, so the correct relation is 2πr=nλ2\pi r = n\lambda.

  • Substituting n=2n = 2 too early and then comparing carelessly with the given options can cause mismatch. First derive the general expression for λ\lambda and then identify the matching option format.

Practice more Dual Nature of Matter & de Broglie Relation questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Dual Nature of Matter & de Broglie Relation

Related questions