The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum…
- $5 \times 10^{-3}$
- $7.3 \times 10^{-3}$
- $3.6 \times 10^{-3}$
- $36.5 \times 10^{-3}$
Solution
where, $\frac{2 \pi k e^2}{h}$ is a constant quantity and $\mathrm{Z}$ is atomic number.
Hence, speed of electron in first excited state
(n = 2) is
$\begin{aligned}
v & =2.188 \times 10^6 \times \frac{1}{2} \quad(Z=1, \text { for hydrogen atom }) \\
& =1.094 \times 10^6 \mathrm{~ms}^{-1}
\end{aligned}$
Ratio of speed of electron in lst excited state to
$\begin{aligned}
\text {the speed of light in vacuum } & =\frac{1.094 \times 10^6}{3 \times 10^8} \\
& =0.364 \times 10^{-2} \\
& =3.6 \times 10^{-3}
\end{aligned}$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)