Atomic Structure

1. Relate de Broglie Wavelength and Bohr Radius

According to Bohr's quantization condition (postulate), the angular momentum of an electron in a stationary orbit is quantized:

$ m v r_n h n \lambda mv h r_n \lambda \lambda = 13.4 \text{ Å} n^{\text{th}} r_n = 0.53n^2 \text{ Å} \pi = 3.14 r_n \lambda 2\pi r_n = n\lambda n n n n 4.0128 n$ is 4.

The wavelength . The Rydberg formula for the wavelength of light absorbed/emitted by a hydrogen-like atom is given by:

For the longest wavelength, the transition is between consecutive levels and :

Substituting the known values:

Simplifying gives:

Calculate :

Computing :

Rounding to the nearest integer:

Step 1: Analysis:

The probability density is |ψ(x,y,z)|², and the probability of finding the particle in a small volume element dx dy dz is:

P = |ψ(x,y,z)|² dx dy dz

Step 2: Evaluating each option:

*(A) ψ(x,y,z) ψ(x,y,z)**: ✗

• Missing the volume element dx dy dz
• This is just the probability density, not the probability

(B) |ψ(x,y,z)|² dx dy dz: ✓

• This is the standard form
• |ψ|² = ψ*ψ is the probability density
• Multiplied by the volume element gives probability

(C) ψ(x,y,z) ψ(x,y,z) dx dy dz*: ✓

• Since |ψ|² = ψ*ψ, this is equivalent to option (B)
• This explicitly shows the probability density as ψ*ψ
• Includes the volume element

*(D) ∫∫∫ dx dy dz ψ(x,y,z) ψ(x,y,z)**: ✗

• This integral over all space gives the normalization condition = 1
• It gives the total probability (which must equal 1 for a normalized wavefunction)
• Not the probability in a specific small volume element

Answer: (B) and (C) are correct.

To find the ratio of the transition energy for the transition in to that in the H atom, we need to consider the Rydberg formula for calculating the energy of the level in hydrogen-like atoms. The formula for the energy levels is given by:

where:

• is the energy of the level.
• is the atomic number of the element.
• is the Rydberg constant for the hydrogen atom.
• is the principal quantum number.

For a transition from the level ( ) to the level ( ), the energy change is given by:

Simplifying this expression gives:

Now, let's calculate the transition energy ratio between and :

• For , , so .
• For , , so .

The ratio of the transition energies is:

Therefore, the correct answer is 4. This shows that the energy of the transition in is four times that in the hydrogen atom.