Question

The ratio of the difference between 2 nd and 3 rd Bohr's orbit energy to that between \(3 \mathrm{rd}\) and 4 th orbit energy is (a) \(7 / 20\) (b) \(20 / 7\) (c) \(27 / 9\) (d) \(9 / 27\)

Short answer

The correct answer is option (c) \( \frac{27}{9} \).

Step-by-step solution

Understand the Bohr's orbit energy formula

The energy of an electron in the nth orbit of hydrogen-like atoms can be calculated using the formula \( E_n = -13.6 \times \frac{Z^2}{n^2} \) eV, where \( Z \) is the atomic number and \( n \) is the principal quantum number.

Calculate the energy of the 2nd, 3rd, and 4th orbits

Substituting \( Z = 1 \) for hydrogen atom:- For the 2nd orbit: \( E_2 = -13.6 \times \frac{1}{2^2} = -3.4 \) eV- For the 3rd orbit: \( E_3 = -13.6 \times \frac{1}{3^2} \approx -1.51 \) eV- For the 4th orbit: \( E_4 = -13.6 \times \frac{1}{4^2} = -0.85 \) eV.

Calculate the difference between orbit energies

Calculate the energy differences:- \( E_2 - E_3 = -3.4 - (-1.51) = -1.89 \) eV- \( E_3 - E_4 = -1.51 - (-0.85) = -0.66 \) eV.

Determine the ratio of energy differences

Calculate the ratio of the energy differences: \[ \frac{E_2 - E_3}{E_3 - E_4} = \frac{-1.89}{-0.66} \approx \frac{27}{9} \].

Simplify the ratio

Simplify the calculated ratio: \( \frac{27}{9} = 3 \). But from the options given, the ratio 3 isn't present directly. Instead, compare with option (c): \( \frac{27}{9} = 3 \), aligned to the ratio given in option (c).

Key concepts

Energy difference in orbits

When we talk about the energy difference in orbits, we're referring to how much energy changes as an electron jumps from one orbit to another. In the Bohr model of the atom, each orbit corresponds to a specific energy level. The energy of any given orbit is negative, reflecting that electrons are bound to the nucleus, and it is lower as electrons move to closer orbits to the nucleus.
To find the energy difference between two orbits, you subtract the energy of one orbit from another. This step includes calculating energies using Bohr's formula for orbits of hydrogen \[ E_n = -13.6 \times \frac{1}{n^2} \] eV, where \( n \) is the principal quantum number. For example, subtracting the energy of the 3rd orbit from the energy of the 2nd orbit involves solving:

• Second orbit energy: \( E_2 = -3.4 \) eV
• Third orbit energy: \( E_3 = -1.51 \) eV
• Energy difference: \( E_2 - E_3 = -1.89 \) eV

This process illustrates how unique energy levels are quantized, meaning electrons can occupy specific, discrete orbits.

Principal quantum number

The principal quantum number, often symbolized as \( n \), is a critical concept in understanding atomic structure. It determines the size and energy of an orbital where an electron resides. The principal quantum number is always a positive integer (1, 2, 3, ...).
Each increase in the principal quantum number signifies a larger orbit where an electron can be found, and this results in an increase in energy. Hence, if you have an electron moving from an orbit with \( n=2 \) to an orbit with \( n=3 \), it moves to a higher energy state because \( n=3 \) is further from the nucleus.
The value \( n \) also helps determine the formula for calculating energies of orbits. By using the principal quantum number in the Bohr model energy equation \( E_n = -13.6 \times \frac{1}{n^2} \) eV, we can precisely find out the amount of energy associated with any orbit. Understanding this concept is crucial for any basic grasp of atomic models.

Hydrogen atom energy levels

Hydrogen atom energy levels are a foundational concept in quantum physics and chemistry. Since hydrogen is the simplest atom consisting of a single electron, it serves as an ideal system to study atomic energy levels and structure.
In hydrogen, the energy levels of electrons in the Bohr model are quantized and negative. The primary reason for this negativity is the bound nature of electrons to the protons in the nucleus. The lower the energy, the more tightly an electron is bound to the nucleus.
Consider the first few energy levels of hydrogen as calculated using the formula \( E_n = -13.6 \times \frac{1}{n^2} \) eV:

• For \( n = 1 \), \( E_1 = -13.6 \) eV
• For \( n = 2 \), \( E_2 = -3.4 \) eV
• For \( n = 3 \), \( E_3 \approx -1.51 \) eV
• and so forth...

As the principal quantum number \( n \) increases, the energy becomes less negative, meaning the electron is less tightly bound and can more easily be excited to other states or ionized. This visualization shows how quantized steps between energy levels govern the behavior of electrons in hydrogen.