Question 14 Marks
Calculate the orbital period in each of these levels.
Answer
View full question & answer→Let periodic orbital time $T_1, T_2$ and $T_3$ are in $n=1,2$ and 3.
For hydrogen atom, radius of electron's orbit
$r_n=\left(0.529 \times 10^{-10}\right) n^2$ meter
So,$\quad$$r_1=0.529 \times 10^{-10}$ meter
$\begin{array}{l}r_2=4 \times 0.529 \times 10^{-10} \text { meter } \\ r_3=9 \times 0.529 \times 10^{-10} \text { meter }\end{array}$
So, orbital speed $T =\frac{2 \pi r}{v}$
$\begin{aligned} T _1 & =\frac{2 \times 3.14 \times 0.529 \times 10^{-10}}{2.19 \times 10^6} \\ =1.52 \times 10^{-16} sec \end{aligned}$
$T _2=\frac{2 \times 3.14 \times 4 \times 0.529 \times 10^{-10}}{1.094 \times 10^6}$
$\approx 12.16 \times 10^{-16} sec$
and$\quad$$\begin{aligned} T _3 =\frac{2 \times 3.14 \times 9 \times 0.529 \times 10^{-10}}{0.729 \times 10^6} \\ =41.01 \times 10^{-16} sec .\end{aligned}$
For hydrogen atom, radius of electron's orbit
$r_n=\left(0.529 \times 10^{-10}\right) n^2$ meter
So,$\quad$$r_1=0.529 \times 10^{-10}$ meter
$\begin{array}{l}r_2=4 \times 0.529 \times 10^{-10} \text { meter } \\ r_3=9 \times 0.529 \times 10^{-10} \text { meter }\end{array}$
So, orbital speed $T =\frac{2 \pi r}{v}$
$\begin{aligned} T _1 & =\frac{2 \times 3.14 \times 0.529 \times 10^{-10}}{2.19 \times 10^6} \\ =1.52 \times 10^{-16} sec \end{aligned}$
$T _2=\frac{2 \times 3.14 \times 4 \times 0.529 \times 10^{-10}}{1.094 \times 10^6}$
$\approx 12.16 \times 10^{-16} sec$
and$\quad$$\begin{aligned} T _3 =\frac{2 \times 3.14 \times 9 \times 0.529 \times 10^{-10}}{0.729 \times 10^6} \\ =41.01 \times 10^{-16} sec .\end{aligned}$