The radius of the second Bohr orbit for the hydrogen atom is: [Given: Planck's const. h=6.6262×10⁻³⁴Js; mass of electron=9.1091×10⁻³¹kg; charge of electron, e=1.60210×10⁻¹⁹C; permittivity of vacuum, ε₀=8.854185×10⁻¹²kg⁻¹m⁻³A²]
4.76 ˚A
0.529 ˚A
2.12 ˚A
1.65 ˚A
Solution:
r=n²h²/4π²me²k r=2²(6.6262×10⁻³⁴Js)²/4×(3.1416)²×9.1091×10⁻³¹kg×(1.60210×10⁻¹⁹C)²×9×10⁹ r=2.12×10⁻¹⁰m r=2.12 ˚A Hence, the option (C) is the correct answer.