Atoms

The correct answer is (D) :
Energy of photon
Electron will excite to
Since
Radius of atom will be

An atom based on classical theory of Rutherford’s model should collapse as the electrons in a continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy. These electrons losing energy should soon fall into a heavy nucleus collapsing the whole atom.
The correct option is (C) : A classical atom based on Rutherford’s model is doomed to collapse.

The Balmer series involves transitions where an electron in a hydrogen atom drops to the n=2 energy level. The wavelengths of the emitted photons are described by the Rydberg formula:

1/λ = R (1/2² - 1/n²)

Where:

• λ is the wavelength of the emitted photon
• R is the Rydberg constant (approximately 1.097 x 10⁷ m^-1)
• n is the principal quantum number of the initial energy level (n > 2)

Hα and Hβ Transitions

• Hα: This corresponds to the transition from n=3 to n=2.
• Hβ: This corresponds to the transition from n=4 to n=2.

Calculating the Wavelengths

1. Wavelength of Hα (λ_α):
   1/λ_α = R (1/2² - 1/3²) = R (1/4 - 1/9) = R (5/36)
   λ_α = 36 / (5R)
2. Wavelength of Hβ (λ_β):
   1/λ_β = R (1/2² - 1/4²) = R (1/4 - 1/16) = R (3/16)
   λ_β = 16 / (3R)

Finding the Ratio λ_α/λ_β

(λ_α / λ_β) = (36 / 5R) / (16 / 3R) = (36 / 5R) * (3R / 16) = (36 * 3) / (5 * 16) = 108 / 80 = 27/20

Determining the Value of x

The problem states that the ratio λ_α/λ_β = x/20. We found that the ratio is 27/20. Therefore, x = 27.

The correct answer is option (A):

Energy of the 2^{nd} orbit: So, only possible transition is from:

The coulombian force ( ) is given by:

The gravitational force ( ) is given by:

The ratio of the forces is:

Therefore, the answer is approximately .

To determine the radius of the fourth stationary orbit relative to the third orbit in Bohr's atom model, we need to understand the relation between the orbit radius and the principal quantum number.

According to Bohr's model, the radius of the -th orbit is given by the formula:

where is the radius of the first stationary orbit, and is the principal quantum number.

Given:
The radius of the third orbit is . Therefore, we have:

From this, we can find :

We want to find the radius of the fourth orbit, :

Substituting the value of , we get:

Thus, the radius of the fourth stationary orbit is .

Therefore, the correct answer is: .

This matches the given correct answer option, confirming our solution is accurate.

To determine the value of for the longest wavelength of radiation in the Paschen series expressed as , we need to analyze the Paschen series of hydrogen. The series is defined for transitions from higher energy levels to . The wavelength of emitted radiation is given by the Rydberg formula: For the longest wavelength, the smallest possible above 3 is 4 ( ): Simplifying gives: Thus, Therefore, Comparing this with , it is clear that , confirming the value falls within the expected range of 144. The longest wavelength of the Paschen series is , validating .

To determine the wavelength of the second member of the Lyman series, we need to understand the Lyman series and apply the Rydberg formula for hydrogen spectral lines. The Lyman series corresponds to electronic transitions where the electrons drop to the n=1 energy level from higher levels (n=2, n=3, etc.).

The Rydberg formula for calculating the wavelength of any transition in a hydrogen atom is given as:

Where:

• is the Rydberg constant ( )
• is the lower energy level (for Lyman series, )
• is the higher energy level

The first member of the Lyman series corresponds to a transition from to . Thus, the wavelength is given by:

The second member of the Lyman series involves a transition from to :

To find the relationship between and , we divide the first equation by the second:

Simplifying, we have:

Thus, the wavelength of the second member is:

Therefore, the correct answer is: .