Question

How many revolutions does an electron in $n=2$ state of a hydrogen atom make before dropping to the ground state? The average life-time of excited state is $10^{-8}$ second. (Given $v_{ n }=2 \cdot 186 \times 10^{6} \times Z / n ms ^{-1}$)

Answer 1

SOLUTION : Velocity in the 2 nd orbit $v_{2}=2 \cdot 186 \times 10^{6} \times 1 / 2 ms ^{-1}$

The distance covered by electron in $10^{-8}$ second

$=1.093 \times 10^{6} \times 10^{-8} m =1.093 \times 10^{-2} m$

Radius of the 2nd orbit $r_{2}=0.529 \times n^{2} Å=4 \times 0.529 \times 10^{-10} m$

Circumference of 2 nd orbit.

$2 \pi r_{2}=2 \times 3 \cdot 1416 \times 4 \times 0.529 \times 10^{-10} m =13 \cdot 29 \times 10^{-10} m$

Number of revolutions $=\frac{1.093 \times 10^{-2}}{13.29 \times 10^{-10}}=0.0822 \times 10^{8}=8.22 \times 10^{6}$.