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QJEEM23Ch20ac: JEE Main 2023 Chemistry - Class 11 JEEM23Ch20ac Solution: If the radius of the first orbit of hydrogen atom is , then de…

If the radius of the first orbit of hydrogen atom is α0\alpha_{0} , then de Broglie's wavelength of electron in 3rd 3^{\text {rd }} orbit is :

Multiple Choice Options

  • A πα36\frac{\pi \alpha^{3}}{6}
  • B 3πα03\pi\alpha_0
  • C 6πα06\pi\alpha_0
  • D πα33\frac{\pi \alpha^{3}}{3}

Step-by-Step Verified Solution

In Bohr's model, the angular momentum of an electron in an orbit is quantized, which can be represented as:

mvr=n⋅h2πmvr = n \cdot \frac{h}{2\pi}

Where:

  • (m) is the mass of the electron,
  • (v) is the velocity of the electron,
  • (r) is the radius of the nth orbit, and
  • (n) is the principal quantum number ((n = 1) for the first orbit, (n = 2) for the second orbit, and so on).

The de Broglie wavelength ( λ\lambda ) is given by:

λ=hmv\lambda = \frac{h}{mv}

For the nth orbit, the radius (r) is n2n^2 times the radius of the first orbit ( r0r_0 , or α0\alpha_0 in your notation):

r=n2⋅α0r = n^2 \cdot \alpha_0

The circumference of the nth orbit is 2πr2\pi r , which can be written as:

2πr=n⋅λ2\pi r = n \cdot \lambda

Substituting the expression for (r):

2π⋅n2⋅α0=n⋅λ2\pi \cdot n^2 \cdot \alpha_0 = n \cdot \lambda

Solving for λ\lambda gives:

λ=2π⋅n⋅α0\lambda = 2\pi \cdot n \cdot \alpha_0

For the 3rd orbit ((n = 3)), this simplifies to:

λ=6π⋅α0\lambda = 6\pi \cdot \alpha_0

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