Discussion Forum

The radius of the second Bohr orbit for hydrogen atom is [Planck's constant (h)= 6.6262 × 10^-34 Js; mass of electron =9.1091 × 10^-31kg ; charge of electron  (e) = 1.60210 × 10^-19 C : permitivity of vacuum (ε₀)=8.854185 × 10^-12 kg^-1 m^-3 A²)

Answer:

Explanation:

Bohr radius (r_n) = $\epsilon_{0}n^{2}h^{2}$

               $r_{n}=\frac{n^{2}h^{2}}{4\pi^{2}me^{2}kZ}$

                     $k=\frac{1}{4\pi \epsilon_{0}}$

      $r_{n}=\frac{n^{2}h^{2}\epsilon_{0}}{\pi^{}me^{2}Z}$

             = $n^{2}\frac{a_{0}}{Z}$

      where m = mass of electron

                e= charge of electron

               h = plank's constant 

               k= Coulomb constant

              $r_{n}= \frac{n^{2}\times 0.53}{Z}\dot{A}$

      Radius of n^th Bohr orbit for H -atoms

                                    $=0.53 n^{2}\dot{A}$

                                                 [ Z= 1 for H-atom]

$\therefore$  Radius of 2^nd Bohr orbit for H-atom

                                 =0.53 × (2)² = $2.12 \dot{A}$