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1800-102-2727For many years, people believed that the sun and the other planets rotated around the earth, placing it at the centre of the solar system. Nicolaus Copernicus proposed that the solar system's centre is the sun, and that the planets orbit it, in 1543. The term for this is the heliocentric theory. Galileo Galilei's observation that the moons Callisto, Europa, Ganymede, and Io are in orbit around Jupiter further supported this theory.
I assume you are aware of the solar system. What relevance does it have to this concept page, though? Calm down; we'll get to that in a second.
In an atom, the nucleus lies at the centre and the electrons orbit it, just like the sun and planets do. Rutherford presented this model of an atom, therefore the name Rutherford's model of an atom. He was unable to demonstrate the location, arrangement, and the reason why the electrons did not fall back onto the nucleus due to electrostatic forces.
You may be thinking, "Isn't this page about Bohr's model? "
For hydrogen or species that are similar to hydrogen (like H, He^{+}, Li^{2+}), Niels Bohr suggested a new model of atoms that addressed some of the features that Rutherford's model was unable to explain. The postulates that served as the foundation for Bohr's theory were based on Planck's quantum theory.
In this concept page, we will get to know more about the Bohr’s model of the atom.
TABLE OF CONTENTS
$mvr=\frac{nh}{2\pi}$ (where n is the orbit number in which the electron is present)
Assume an electron having mass m and charge e revolving in orbit around the nucleus of an atom.
The atomic number of atom = Z = number of proton
Charge of nucleus = Ze (where, e = charge of an electron)
Radius of orbit = r A^{0}
Tangential velocity of the electron in orbit = v m sec^{-1}.
For a stable orbit
Centrifugal force on electron = Electrostatic force of attraction between the nucleus and electron
$\Rightarrow $ $\frac{m{v}^{2}}{r}$ = $\frac{Kz{e}^{2}}{{r}^{2}}$
$\Rightarrow $${v}^{2}$ =$\frac{Kz{e}^{2}}{{mr}^{}}$ ………………… (i)
From the postulates of Bohr’s model,
$\Rightarrow $ mvr = $\frac{nh}{2\pi}$
$\Rightarrow $ v = $\frac{nh}{2mr\pi}$ ………………… (ii)
Substituting (ii) in (i),
$\Rightarrow $ $\frac{{n}^{2}{h}^{2}}{4{m}^{2}{r}^{2}{\pi}^{2}}$ = $\frac{Kz{e}^{2}}{{mr}^{}}$
r = $\frac{{\mathit{n}}^{2}{\mathit{h}}^{2}}{4{\mathit{\pi}}^{2}\mathit{m}\mathit{K}\mathit{Z}{\mathit{e}}^{2}}$ ………………… (iii)
Where, n= energy level
h = Plank’s constant
m = mass of electron
K= Coulomb's constant = $\frac{1}{4\pi {\epsilon}_{0}}$ =$9\times {10}^{9}N{m}^{2}{C}^{-2}$
Z = atomic number of element
e = charge on single electron
${\epsilon}_{0}$ = The permittivity of free space
On substituting the values of e, h,m, K and , the radius of n^{th} Bohr orbit can be calculated.
${\mathit{r}}_{\mathit{n}}=\mathit{}0.529\frac{{\mathit{n}}^{2}}{\mathit{Z}}{\mathit{A}}^{0}$
We know that,
r = $\frac{{\mathit{n}}^{2}{\mathit{h}}^{2}}{4{\mathit{\pi}}^{2}\mathit{m}\mathit{K}\mathit{Z}{\mathit{e}}^{2}}$
mvr= $\frac{nh}{2\pi}$
v= $\frac{nh}{2mr\pi}$
Substituting the value of r in v,
v = $\frac{{2\mathit{\pi}\mathit{Z}{\mathit{e}}^{2}\mathit{K}}^{}}{\mathit{n}\mathit{h}}$........................ (iv)
Where, n = energy level
h = Plank’s constant
m= mass of electron
K = coulomb's constant
Z = atomic number of element
e = charge on single electron
Substituting the values of e, h,m, K and , the velocity of an electron in Bohr’s orbit can be calculated.
${\mathit{v}}_{\mathit{n}}\mathit{}=\mathit{}2.18\mathit{}\times {10}^{6}\frac{\mathit{Z}}{\mathit{n}}$ m sec^{-1}
Time period of revolution (T) is defined as the time required by an electron to complete one revolution in its orbit. The time period of revolution can be calculated by taking the ratio of the radius of Bohr’s orbit and the velocity of an electron in an orbit.
Mathematically,
Time period (T) ∝ $\frac{Ratiooftheboh{r}^{\text{'}}sradius}{Velocityofanelectroninanorbit}$ ........................ (v)
From equation (iii), the radius of Bohr’s orbit $\propto \frac{{n}^{2}}{Z}$
From equation (iv), the velocity of an electron in an orbit $\propto \frac{Z}{n}$
Where, n = energy level
Z = atomic number of the element
Substituting the values of the radius of Bohr’s orbit and the velocity of an electron,
Time period (T) ∝ $\frac{\frac{{n}^{2}}{Z}}{\frac{Z}{n}}$
Time period (T) $\propto \frac{{n}^{3}}{{Z}^{2}}$ ........................ (vi)
Frequency of revolution is defined as the number of revolution completed by an electron in one sec. It is calculated by taking the reciprocal of the time period of the revolution.
Mathematically,
Frequency of the revolution (f) = $\frac{1}{Timeperiod\left(T\right)}$
From equation (vi),
Frequency of the revolution (f) ∝ $\frac{{Z}^{2}}{{n}^{3}}$ ........................ (vii)
We know that the total energy of an electron moving in an orbit (T.E) = Kinetic energy (K.E) + Potential energy (P.E)
Kinetic energy (K.E) = $\frac{1}{2}m{V}^{2}$
Potential energy (P.E) = -$\frac{KZ{e}^{2}}{r}$
(nuclear charge = Ze and charge of electron = - e)
T.E = $\frac{1}{2}m{V}^{2}$ - $\frac{KZ{e}^{2}}{r}$ ........................ (viii)
Substituting (i) in (vii),
T.E = $\frac{KZ{e}^{2}}{2r}$ - $\frac{KZ{e}^{2}}{r}$ = - $\frac{KZ{e}^{2}}{r}$ ...................... (ix)
Substituting the value r in (viii),
E = - $\frac{{2{\mathit{\pi}}^{2}\mathit{m}{\mathit{e}}^{4}{\mathit{K}}^{2}}^{}}{{\mathit{h}}^{2}}\left(\frac{{\mathit{Z}}^{2}}{{\mathit{n}}^{2}}\right)$
Where, 'n' represent the orbit in which electron revolves
h = Plank’s constant
m = mass of electron
K = Coulomb's constant
Z = atomic number of element
e = charge on single electron
Substituting the values of e, h,m, K and , the energy of an electron in Bohr’s orbit can be calculated.
${\mathit{E}}_{\mathit{n}}\mathit{}=\mathit{}-13.6\frac{{\mathit{Z}}^{2}}{{\mathit{n}}^{2}}\mathit{e}\mathit{V}\mathit{}\mathit{a}\mathit{t}\mathit{o}{\mathit{m}}^{-1}$
Note: T.E = $\frac{\mathit{P}.\mathit{E}}{2}$ and T.E = - K.E
Conclusions from the equation of energy
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Q1. Calculate the radius ratio for 3^{rd} orbit of Li^{2+} ion & 4th orbit of Be^{3+} ion.
A. $\frac{3}{4}$
B. $\frac{4}{7}$
C. $\frac{9}{8}$
D. $\frac{5}{2}$
Answer: A)
Solution: We know that, r=0.529 $\frac{{\mathit{n}}^{2}}{\mathit{Z}}$A^{0}
${\mathit{r}}_{\mathit{L}{\mathit{i}}^{2+}}\mathit{}=\mathit{}0.529\frac{(3{)}^{2}}{3}{\mathit{A}}^{0}$ ………………… (i)
${\mathit{r}}_{\mathit{B}{\mathit{e}}^{3+}}\mathit{}=0.529\frac{(4{)}^{2}}{4}{\mathit{A}}^{0}$ ………………… (ii)
Dividing (i) by (ii),
$\frac{{\mathit{r}}_{\mathit{L}{\mathit{i}}^{2+}}}{{\mathit{r}}_{\mathit{B}{\mathit{e}}^{3+}}\mathit{}}$ = $\frac{3}{4}$
So, option A is the correct answer.
Q2. Find the orbit number of Li^{2+} ion in which the electron is revolving with a speed of$1.635\mathit{}\times \mathit{}{10}^{6}\mathit{}\mathit{m}\mathit{}\mathit{s}\mathit{e}{\mathit{c}}^{-1}$.
A. 4
B. 3
C. 2
D. 1
Answer: A)
Solution: We know that, $\mathit{V}\mathit{}=\mathit{}2.18\mathit{}\times {10}^{6}\frac{\mathit{Z}}{\mathit{n}}\mathit{m}\mathit{}\mathit{s}\mathit{e}{\mathit{c}}^{-1}$
$1.635\mathit{}\times \mathit{}{10}^{6}\mathit{}\mathit{m}\mathit{}\mathit{s}\mathit{e}{\mathit{c}}^{-1}\mathit{}\mathit{}\mathit{}=\mathit{}2.18\mathit{}\times {10}^{6}\frac{3}{\mathit{n}}\mathit{m}\mathit{}\mathit{s}\mathit{e}{\mathit{c}}^{-1}$
n= $\frac{2.18\mathit{}\times \mathit{}3}{1.635}$ =4
So, option A is the correct answer.
Q3. Bohr’s model can be applied for
A. Hydrogen
B. Helium
C. Boron
D. None of the above
Answer: A)
Solution: Helium has 2 electrons and boron has 5 electrons. Bohr’s model is only applicable for single-electron species like H, He^{+} etc.
So, option A is the correct answer.
Q4. The approximate radius of 1st Bohr’s orbit of hydrogen atom is
A. 1 Å
B. 0.229 Å
C. 0.345 Å
D. 0.529 Å
Answer: D)
Solution: We know that, $\mathit{r}\mathit{}=\mathit{}0.529\frac{{\mathit{n}}^{2}}{\mathit{Z}}{\mathit{A}}^{0}$
For a hydrogen atom, Z =1 and n = 1
$r=0.529\frac{{1}^{2}}{1}{A}^{0}=0.529{A}^{0}$
So, option D is the correct answer.
Q1. Why Bohr’s atomic model is only valid for hydrogen or hydrogen-like species?
Answer: Bohr’s atomic model is only valid for hydrogen or hydrogen-like species because only Coulombic interactions between one proton and one electron are considered in Bohr's model for a hydrogen atom. It cannot be extended to include other atomic species with more than one electron. Because, in this case, the interactions between electrons of the same species occur in addition to the interactions between the nucleus and the electron. Bohr was unable to solve this problem, which is now successfully explained by later developed quantum mechanics. However, Bohr's model can be successfully applied to hydrogen-like species such as He^{+}, Li^{2+} and so on. Therefore it applies only to single electron species.
Q2. Are all postulates of Bohr’s atomic model correct according to the advancements in scientific research?
Answer: No, the concept of a fixed path (orbit) is not valid after future research. Bohr’s orbit does not follow Heisenberg's uncertainty principle.
Q3. Why doesn’t a flawless model for atomic structure exist?
Answer: Since the atoms are too small to even view through a microscope, no one can tell what the shape of an atom is. The theories and models of atoms mentioned have been proposed after years of research and are true to date.
Q4. How does Bohr’s atomic model explain the stability of an atom?
Answer:According to Bohr’s atomic model, electrons revolve in a fixed path known as the orbit with a particular amount of energy. The value of energy does not change until and unless it absorbs or emits energy. This model also explains the stability of atoms as an electron cannot lose more energy than it has in the smallest orbit, n = 1. The model also explains the Balmer formula for hydrogen spectral lines.