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AP EAMCET201726 Apr 2017Morning ShiftPhysicsAtomic PhysicsActual

The radius of orbit of an electron and the speed of electron in the ground state of hydrogen atom are 5.5 10⁻¹¹ ~m and 4 10^6 ~ms ⁻¹ respectively. Then, the orbital period of this electron in the first excited state will be ......... .

Options

  1. A6.90810⁻¹⁶ ~s
  2. B9.608 10⁻¹⁶ ~s
  3. C7.806 10⁻¹⁶ ~s
  4. D8.9068 10⁻¹⁶ ~s

Correct answer

A. 6.90810⁻¹⁶ ~s

Step-by-step solution

r₁=5.5 10⁻¹¹ ~m v₁=4 10^6 ~m / s As r_n n^2 So, r₂=5.5 10⁻¹¹ 2^2=22.0 10⁻¹¹ ~m / s As v_n 1 n So, v₂= 4 10^6 2 =2 10^6 ~m / s The time period of revolution aligned & T= 2 r₂ v₂ = 2 3.14 22.0 10⁻¹¹ 2 10^6 & =6.908 10⁻¹⁶ ~s aligned

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