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# Bohr Hydrogen Atom Assignment Help

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Atomic Physics - Bohr Hydrogen Atom

**Bohr Hydrogen Atom**

In view of the limitations of Rutherford atom model, Niels Bohr came to the conclusion that classical mechanics and electromagnetism could not be applied to the processes on the atomic scale. Bohr cleverly combined classical ideas and early quantum concepts (given by Planck) to give what is known as Bohr model of hydrogen atom. Following are the three postulates of this model:

1. Every atom consists of a central core called nucleus. In which entire positive charge and almost entire mass of the atom are concentrated. A suitable number of electrons (having as much negative charge as the positive charge on the nucleus) revolve around the nucleus in circular orbits. The centripetal force required for revolution is provided by the electrostatic force of attraction between the electron and the nucleus.

This postulate stresses that an electron in an atom can revolve in certain stable orbits only without the emission of radiant energy. Thus, each atomic exist only in certain definite stable states with definite total energy. These are called stationary states of the atom.

If **m** is the mass of electron moving with a velocity v in a circular orbit of radius r, then the necessary centripetal force is **F = mv2/r**

Also, the electrostatic force of attraction between the nucleus of charge** (+Ze)** and electron of charge** (-e)**is

**F = 1/4∏?**_{0} [(Z_{e}) (e)/r^{2}] = KZe^{2}/r^{2}

Where **K = 1/4∏?**_{0}

Therefore, **mv**^{2}/r = KZ*e^{2}/r^{2} (1)

**2.** Bohr’s second postulate defines these stable orbits.

According to Bohr, electron can revolve only in certain discrete non radiating orbits, called stationary orbits, for which total angular momentum of revolving electron is an integral multiple of **h/2∏**, where h is Planck’s constant.

Thus the angular momentum of the orbiting electron is quantized.

As angular momentum of electron **= mvr,**

Therefore, for any permitted (stationary) orbit

**mvr = nh/2∏ (2)**

where n is any positive integer,** 1, 2, 3…….**

It is called principle quantum number. The quantum condition **(10) **limits the number of allowed orbits. The electron, while revolving in such orbits, shall not lose energy i.e. its energy would stay constant.

**3. **The emission/absorption of energy occurs only when an electron jumps from one of its specified non-radiating orbit to another. The difference in the total energy of electron in the two permitted orbits is absorbed when the electron jumps from inner to the outer orbit, and emitted when electron jumps from outer to the inner orbit.

If **E**_{1} is total energy of electron in an inner stationary orbit and **E**_{2} is its total energy in an outer stationary orbit, then frequency v of radiation emitted on jumping from outer to inner orbit is given by

**hv = E**_{2} – E_{1 } (3)

**(a) **Radii of Bohr’s stationary orbits

**From (2), mvr = nh/2∏,**

**Or v = nh/2∏mr**

Put in (1),** m/r × n**^{2}h^{2}/4∏2m^{2}r^{2} = KZe^{2}/r^{2}

Or r = n^{2}h^{2}/4∏2m KZe^{2}

For hydrogen atom, Z = 1

Therefore,** r = n**^{2}h^{2}/4∏2m Ke^{2} (4)

This shows that **r ∝ n**^{2 }

i.e. radii of stationary orbits are in the ratio **12: 22: 32: **and so on i.e. **1:4:9 **clearly, the stationary orbits are not equally spaced.

Putting **h = 6.6 × 10**^{-34} joule-sec.

**M = 9.1 × 10-31 kg.**

K = 9 × 109 Nm2 C-2,

E = 1.6 × 10-19 C

We get from eqn. (4)

**R = n**^{2} × 5.29 × 10^{-11} m

For example, size of **1st orbit (n = 1) **of hydrogen atom is

**R = 1 × 5.29 × 10**^{-11} m = 0.529 λ .

**(b) **Velocity of electrons in Bohr’s stationary orbit

From (1), **r = KZe**^{2}/mv^{2}

From (2) **r = nh/2∏ mv**

Or** v = 2∏ KZe2/nh (5)**

For hydrogen atom,** Z = 1**

Therefore, **v = Ke**^{2}/nh (6)

Calculation show that in first orbit** (n = 1) **of hydrogen atom, orbital velocity of electrons is** 2.2 × 10**^{6 }m/swhich is roughly **1/137** of the velocity of light in vacuum.

Further, the orbital velocity of electron in outer orbits is smaller as compared to its value in the inner orbits.

(c) Frequency of electrons in Bohr’s stationary orbit

It is the number of revolutions completed per second by the electron in a stationary orbit, around the nucleus. It is represented by v

From **v = rw = r (2∏ v)**

**V = v/2∏r = 2∏ KZe**^{2}/nh × 2∏r = KZ e^{2}/nhr

i.e.** v = KZ e**^{2}/nhr using (5)

In the first orbit of hydrogen atom,

**N = 1, r =0.53 × 10**^{-10} m

Using **K = 9 × 10**^{9} Nm^{2} C^{-2},

**Z = 1, h = 6.6 × 10**^{-34 }Js

**E = 1.6 × 10**^{-19 }C, we get

**V = 9 × 10**^{9} × 1 × (1.6 × 10^{-19})2/ (6.6 × 10^{-34}) × (0.53 × 10^{-10})

**V = (6.6 × 10**^{-34}) × (0.53 × 10^{-10})

V = 6.57 × 10^{15} rps.

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**Bohr Hydrogen Atom**

**Every atom consists of a central core called nucleus. In which entire positive charge and almost entire mass of the atom are concentrated. A suitable number of electrons (having as much negative charge as the positive charge on the nucleus) revolve around the nucleus in circular orbits. The centripetal force required for revolution is provided by the electrostatic force of attraction between the electron and the nucleus.**

1.

1.

This postulate stresses that an electron in an atom can revolve in certain stable orbits only without the emission of radiant energy. Thus, each atomic exist only in certain definite stable states with definite total energy. These are called stationary states of the atom.

If

**m**is the mass of electron moving with a velocity v in a circular orbit of radius r, then the necessary centripetal force is

**F = mv2/r**

Also, the electrostatic force of attraction between the nucleus of charge

**(+Ze)**and electron of charge

**(-e)**is

**F = 1/4∏?**

_{0}[(Z_{e}) (e)/r^{2}] = KZe^{2}/r^{2}

Where

**K = 1/4∏?**

_{0}

Therefore,

**mv**

^{2}/r = KZ*e^{2}/r^{2}(1)**2.**Bohr’s second postulate defines these stable orbits.

According to Bohr, electron can revolve only in certain discrete non radiating orbits, called stationary orbits, for which total angular momentum of revolving electron is an integral multiple of

**h/2∏**, where h is Planck’s constant.

Thus the angular momentum of the orbiting electron is quantized.

As angular momentum of electron

**= mvr,**

Therefore, for any permitted (stationary) orbit

**mvr = nh/2∏ (2)**

where n is any positive integer,

**1, 2, 3…….**

It is called principle quantum number. The quantum condition

**(10)**limits the number of allowed orbits. The electron, while revolving in such orbits, shall not lose energy i.e. its energy would stay constant.

**3.**The emission/absorption of energy occurs only when an electron jumps from one of its specified non-radiating orbit to another. The difference in the total energy of electron in the two permitted orbits is absorbed when the electron jumps from inner to the outer orbit, and emitted when electron jumps from outer to the inner orbit.

If

**E**

_{1}is total energy of electron in an inner stationary orbit and

**E**

_{2}is its total energy in an outer stationary orbit, then frequency v of radiation emitted on jumping from outer to inner orbit is given by

**hv = E**

_{2}– E_{1 }(3)**(a)**Radii of Bohr’s stationary orbits

**From (2), mvr = nh/2∏,**

**Or v = nh/2∏mr**

Put in (1),

**m/r × n**

^{2}h^{2}/4∏2m^{2}r^{2}= KZe^{2}/r^{2}

Or r = n

Or r = n

^{2}h^{2}/4∏2m KZe^{2}

For hydrogen atom, Z = 1

Therefore,

**r = n**

^{2}h^{2}/4∏2m Ke^{2}(4)This shows that

**r ∝ n**

^{2 }

i.e. radii of stationary orbits are in the ratio

**12: 22: 32:**and so on i.e.

**1:4:9**clearly, the stationary orbits are not equally spaced.

Putting

**h = 6.6 × 10**

^{-34}joule-sec.**M = 9.1 × 10-31 kg.**

K = 9 × 109 Nm2 C-2,

E = 1.6 × 10-19 C

K = 9 × 109 Nm2 C-2,

E = 1.6 × 10-19 C

We get from eqn. (4)

**R = n**

^{2}× 5.29 × 10^{-11}mFor example, size of

**1st orbit (n = 1)**of hydrogen atom is

**R = 1 × 5.29 × 10**λ .

^{-11}m = 0.529**(b)**Velocity of electrons in Bohr’s stationary orbit

From (1),

**r = KZe**

^{2}/mv^{2}

From (2)

**r = nh/2∏ mv**

Or

**v = 2∏ KZe2/nh (5)**

For hydrogen atom,

**Z = 1**

Therefore,

**v = Ke**

^{2}/nh (6)Calculation show that in first orbit

**(n = 1)**of hydrogen atom, orbital velocity of electrons is

**2.2 × 10**which is roughly

^{6 }m/s**1/137**of the velocity of light in vacuum.

Further, the orbital velocity of electron in outer orbits is smaller as compared to its value in the inner orbits.

(c) Frequency of electrons in Bohr’s stationary orbit

It is the number of revolutions completed per second by the electron in a stationary orbit, around the nucleus. It is represented by v

From

**v = rw = r (2∏ v)**

**V = v/2∏r = 2∏ KZe**

^{2}/nh × 2∏r = KZ e^{2}/nhri.e.

**v = KZ e**

^{2}/nhr using (5)In the first orbit of hydrogen atom,

**N = 1, r =0.53 × 10**

^{-10}mUsing

**K = 9 × 10**

^{9}Nm^{2}C^{-2},**Z = 1, h = 6.6 × 10**

^{-34 }Js**E = 1.6 × 10**, we get

^{-19 }C**V = 9 × 10**

^{9}× 1 × (1.6 × 10^{-19})2/ (6.6 × 10^{-34}) × (0.53 × 10^{-10})**V = (6.6 × 10**

V = 6.57 × 10

^{-34}) × (0.53 × 10^{-10})V = 6.57 × 10

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