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The angular speed of electron in nth orbit of hydrogen atom is 1) directly proportional to n² 2)directly proportional to n 3)inversely proportional to n³ 4)inversely proportional to n?
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The angular speed of electron in nth orbit of hydrogen atom is 1) dire...
Answer:

Introduction:
The hydrogen atom is the simplest atom, consisting of one proton and one electron. The electron moves around the nucleus in circular orbits. The angular speed of the electron in the nth orbit of hydrogen atom is proportional to the principle quantum number (n).

Explanation:

Angular Speed:
Angular speed is defined as the rate of change of angle with respect to time. In the case of hydrogen atom, the electron moves in a circular orbit around the nucleus. The angular speed of the electron is given by:

ω = v/r

where ω is the angular speed, v is the linear speed of the electron, and r is the radius of the orbit.

Bohr's Model:
Bohr's model of the hydrogen atom is based on the following assumptions:

1. The electron moves in circular orbits around the nucleus.
2. The electron can only occupy certain discrete orbits, and each orbit has a specific energy level.
3. The electron emits or absorbs energy when it moves from one orbit to another.

The energy of the electron in the nth orbit is given by:

E = -13.6/n² eV

where E is the energy of the electron, and n is the principle quantum number.

Derivation:
Using the above equations, we can derive the expression for the angular speed of the electron in the nth orbit of hydrogen atom:

ω = v/r

v = 2πr/T

where T is the time period of the electron in the nth orbit.

T = 2πr/v

Substituting the value of T in the expression for v, we get:

v = (2πr)/(2πr/v) = (2πr)²/T

The energy of the electron in the nth orbit is given by:

E = -13.6/n² eV

The angular momentum of the electron is given by:

L = mvr

where m is the mass of the electron.

Using the above equations, we can derive the expression for the angular speed of the electron in the nth orbit of hydrogen atom:

ω = L/mr² = (nh/2πmr²)

where h is the Planck's constant.

Substituting the value of v in the expression for ω, we get:

ω = (nh/2πmr²) = (n²h²/4π²mr⁴) x (2πr)²/(-13.6/n²)

ω = (-n³h²/4π²mr⁴) x (13.6 eV/n²)

ω = (-13.6n³/4π²mre²) x (1/n²)

where e is the charge on the electron.

Conclusion:
From the above derivation, we can conclude that the angular speed of the electron in the nth orbit of hydrogen atom is inversely proportional to n³. Therefore, option (3) is correct: inversely proportional to n³.
Community Answer
The angular speed of electron in nth orbit of hydrogen atom is 1) dire...
V= z/ n
r= n2/ z
angular speed ( w) = v/ r
z2/n3
so inversly proportional to n3
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