Search any question & find its solution
Question:
Answered & Verified by Expert
Based on the Bohr's theory of hydrogen atom, the speed of the electron, energy of the electron and the radius of its orbit varies with the principal quantum number, respectively as
Options:
Solution:
2499 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{n}: \frac{1}{n^2}: n^2$
According to Bohr's theory of hydrogen at
(i) The speed of electron in the $n$th orbit,
$$
\begin{aligned}
v_n & =\frac{1}{n} \frac{e^2}{4 \pi \varepsilon_0(h / 2 \pi)} \\
\text { or, } \quad v_n & \propto \frac{1}{n}
\end{aligned}
$$
(ii) The energy of electron in the $n$th orbit,
$$
\begin{aligned}
E_n & =\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2} \\
\text { or, } E_n & \propto \frac{1}{n^2}
\end{aligned}
$$
(iii) The radius of the electron in $n$th orbit,
$$
r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e} \quad \text { or } \quad r_n \propto n^2
$$
So, the correct ratio is $1 / n: 1 / n^2: n^2$.
(i) The speed of electron in the $n$th orbit,
$$
\begin{aligned}
v_n & =\frac{1}{n} \frac{e^2}{4 \pi \varepsilon_0(h / 2 \pi)} \\
\text { or, } \quad v_n & \propto \frac{1}{n}
\end{aligned}
$$
(ii) The energy of electron in the $n$th orbit,
$$
\begin{aligned}
E_n & =\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2} \\
\text { or, } E_n & \propto \frac{1}{n^2}
\end{aligned}
$$
(iii) The radius of the electron in $n$th orbit,
$$
r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e} \quad \text { or } \quad r_n \propto n^2
$$
So, the correct ratio is $1 / n: 1 / n^2: n^2$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.