Q.
Among the following statements, choose the code for the ones that are correct regarding Bohr's model for hydrogen atom.
$\left(\right.I\left.\right)$ Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus
$\left(\right.II\left.\right)$ Radii of allowed orbits of electron are proportional to the principal quantum number
$\left(\right.III\left.\right)$ Frequency with which electrons orbits around the nucleus in discrete orbits is inversely proportional to principal quantum number's cube.
$\left(\right.IV\left.\right)$ Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits
NTA AbhyasNTA Abhyas 2022
Solution:
Speed of electron in $n^{t h}$ orbit = $2.18\times 10^{6}\times \frac{Z}{n}$
For hydrogen atom, $Z=1$
Speed of electron in $n^{t h}$ 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 $n^{t h}$ orbit = $0.529\frac{n^{2}}{Z}\overset{o}{A}$
Radii of an electron for hydrogen atom in $n^{t h}$ orbit = $0.529\frac{n^{2}}{1}\overset{o}{A}$
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 = $\frac{K Z e^{2}}{r^{2}}$
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.
For hydrogen atom, $Z=1$
Speed of electron in $n^{t h}$ 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 $n^{t h}$ orbit = $0.529\frac{n^{2}}{Z}\overset{o}{A}$
Radii of an electron for hydrogen atom in $n^{t h}$ orbit = $0.529\frac{n^{2}}{1}\overset{o}{A}$
Radii is directly proportional to $n^{2}$ , not to $n$ .
Hence, $II$ is incorrect.
$T$ 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 = $\frac{K Z e^{2}}{r^{2}}$
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.