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Atoms and Nuclei question

2019 · 12 Apr · Shift 2 · Q57
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Atoms and Nuclei question

2019 · 12 Apr · Shift 2 · Q57

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 Ao\mathop A\limits^oAo​). The de-Broglie wavelength of this electron is :
  1. A
    6.6 Ao\mathop A\limits^oAo​
  2. B
    3.5 Ao\mathop A\limits^oAo​
  3. C
    9.7 Ao\mathop A\limits^oAo​
  4. D
    12.9 Ao\mathop A\limits^oAo​
View written solutionFree

Correct answer: C

  1. Identify the orbit

    The atom is in its second excited state.

    • Ground state ⇒n=1\Rightarrow n=1⇒n=1
    • First excited state ⇒n=2\Rightarrow n=2⇒n=2
    • Second excited state ⇒n=3\Rightarrow n=3⇒n=3

    So, the electron is in the orbit with principal quantum number n=3n=3n=3

  2. Use Bohr’s standing wave condition

    For an electron in the nthn^{\text{th}}nth orbit, 2πrn=nλ2\pi r_n = n\lambda2πrn​=nλ

    Hence, λ=2πrnn\lambda = \frac{2\pi r_n}{n}λ=n2πrn​​

  3. Substitute the given values

    Given radius, rn=4.65 A˚r_n = 4.65\,\mathring{A}rn​=4.65A˚ and n=3n=3n=3.

    Therefore, λ=2π×4.653 A˚\lambda = \frac{2\pi \times 4.65}{3}\,\mathring{A}λ=32π×4.65​A˚

  4. Calculate

    2π×4.65≈29.222\pi \times 4.65 \approx 29.222π×4.65≈29.22

    So, λ≈29.223=9.74 A˚\lambda \approx \frac{29.22}{3} = 9.74\,\mathring{A}λ≈329.22​=9.74A˚

    Therefore, λ≈9.7 A˚\lambda \approx 9.7\,\mathring{A}λ≈9.7A˚

  5. Check options

    • A: 6.6 A˚6.6\,\mathring{A}6.6A˚
    • B: 3.5 A˚3.5\,\mathring{A}3.5A˚
    • C: 9.7 A˚9.7\,\mathring{A}9.7A˚
    • D: 12.9 A˚12.9\,\mathring{A}12.9A˚

    The correct option is: C\boxed{\text{C}}C​

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