Question

Among the following statements, choose the code for the ones that are correct regarding Bohr's model for hydrogen atom.
$(I)$ Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus
$(II)$ Radii of allowed orbits of electron are proportional to the principal quantum number
$(III)$ Frequency with which electrons orbits around the nucleus in discrete orbits is inversely proportional to principal quantum number's cube.
$(IV)$ Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits

• A. $I$ and $III$
• B. $II$ and $IV$
• C. $I,II$ and $III$
• D. $II,III$ and $IV$

Answer 1

Answer: (A)

Speed of electron in $n^{th}$ orbit = $2.18\times 10^6\times \frac{Z}{n}$
For hydrogen atom, $Z=1$
Speed of electron in $n^{th}$ orbit for hydrogen atom = $\frac{2.18\times 10^6}{n}$
So, as the value of $n$ increases, orbiting speed of electron decreases.
Hence, $I$ is correct.
Radii of an electron in $n^{th}$ orbit = $0.529\frac{n^2}{Z}\stackrel{o}{A}$
Radii of an electron for hydrogen atom in $n^{th}$ orbit = $0.529\frac{n^2}{1}\stackrel{o}{A}$
Radii is directly proportional to $n^2$ , not to $n$ .
Hence, $II$ is incorrect.

Time taken by an electron to complete one revolution = $1.534\times 10^{-10}\times \frac{n^3}{Z^2}s$
$T$ is proportional to $\frac{n^3}{Z^2}$
We know, frequency with which electrons orbit in nucleus $f$ = $\frac{1}{T}$
So, $f$ is inversely proportional to $n^3$ for hydrogen atom
Hence, $III$ is correct.
Binding force with which electrons is bound to nucleus = $\frac{KZ{e}^2}{r^2}$
In hydrogen atom, binding force $=\frac{K{e}^2}{r^2}$
And, we know, as we shift to outer orbit, $r$ increases. So, binding force decreases.
Hence, $IV$ is incorrect.