Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

The frequency of radiation emitted when the electron falls from \(\mathrm{n}=4\) to \(\mathrm{n}=1\) in a hydrogen atom will be (Given ionization energy of \(\mathrm{H}=2.18 \times 10^{18} \mathrm{~J}\) atom \(^{-1}\) and \(\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}\) ) (a) \(1.54 \times 10^{15} \mathrm{~s}^{-1}\) (b) \(1.03 \times 10^{15} \mathrm{~s}^{-1}\) (c) \(3.08 \times 10^{15} \mathrm{~s}^{-1}\) (d) \(2.00 \times 10^{15} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The frequency of the emitted radiation is \( 3.08 \times 10^{15} \) s\(^{-1}\), which corresponds to option (c).

Step 1: Understanding the Rydberg Formula for Energy Transition

The energy of an electron transition between two levels in a hydrogen atom is given by the Rydberg formula: \( \Delta E = -2.18 \times 10^{-18}(\frac{1}{n_2^2} - \frac{1}{n_1^2}) \) J, where \( n_1 \) and \( n_2 \) are the principal quantum numbers for the initial and final states, respectively. Here, \( n_1 = 4 \) and \( n_2 = 1 \).

Step 2: Calculate the Energy Difference

Use the formula \( \Delta E = -2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \) J. This evaluates to \( \Delta E = -2.18 \times 10^{-18} \left( 1 - \frac{1}{16} \right) = -2.18 \times 10^{-18} \times \frac{15}{16} \).

Step 3: Simplify the Energy Expression

Simplify the expression: \( \Delta E = -2.18 \times 10^{-18} \times \frac{15}{16} = -2.04375 \times 10^{-18} \) J. Here, the negative sign indicates a loss of energy as the electron moves to a lower energy level.

Step 4: Relation of Energy to Frequency

To find the frequency \( u \) of the emitted radiation, use the equation \( \Delta E = h u \). Rearrange to solve for frequency: \( u = \frac{\Delta E}{h} \).

Step 5: Calculate the Frequency

Substitute \( \Delta E = 2.04375 \times 10^{-18} \) J and \( h = 6.625 \times 10^{-34} \) Js into the formula to get \( u = \frac{2.04375 \times 10^{-18}}{6.625 \times 10^{-34}} \), which evaluates to \( u \approx 3.08 \times 10^{15} \text{s}^{-1} \).

Step 6: Match the Frequency to the Given Options

Compare the calculated frequency \( 3.08 \times 10^{15}\, \text{s}^{-1} \) with the provided options. The correct match is option (c) \( 3.08 \times 10^{15} \) s\(^{-1}\).

Key Concepts

Rydberg Formula

When it comes to understanding how electrons transition between energy levels in a hydrogen atom, the Rydberg Formula is a central tool. This formula allows us to calculate the energy difference between two energy levels given the principal quantum numbers, which are denoted as \(n_1\) and \(n_2\). The formula is expressed as: \[ \Delta E = -2.18 \times 10^{-18} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] where \(\Delta E\) is the change in energy in joules, \(n_1\) is the initial energy level, and \(n_2\) is the final energy level. For our exercise, these values are 4 and 1, respectively.

• The negative sign indicates a release of energy which is customary when moving to a lower energy level.
• The constant \(-2.18 \times 10^{-18} \) J is specific to hydrogen and signifies the ionization energy for moving from the ground state to an infinite state.

Understanding and applying the Rydberg Formula is essential for calculating transition energies in physics and chemistry, particularly in the case of hydrogen atom experiments.

Energy Levels

Energy levels in atoms are quantized, meaning that electrons occupy defined energy states rather than any continuous range. In the hydrogen atom, these energy levels are determined by the principal quantum number \(n\), where \(n\) is a positive integer.

• The hydrogen atom's energy levels decrease with increasing values of \(n\), approaching zero as \(n\) approaches infinity, which represents the free electron situation or ionization.
• When an electron transitions between energy levels, it either absorbs or emits energy, depending on whether it moves to a higher or lower energy state.

In our exercise, the electron transitions from \(n=4\) to \(n=1\), a large energy difference. This transition falls into the ultraviolet range, as the frequency turns out to be quite high.

Frequency Calculation

Frequency is directly related to the energy change during an electron transition. According to quantum mechanics, the equation \( \Delta E = h u \) connects the energy difference \(\Delta E\) with the frequency of emitted radiation \(u\). Here, \(h\) is Planck's constant, approximately \(6.625 \times 10^{-34} \) Js. To find the frequency \(u\), the equation is rearranged to:\[u = \frac{\Delta E}{h} \]

• This formula tells us that the frequency of radiation emitted or absorbed is proportional to the energy change itself.
• In our exercise, after calculating every necessary component, frequency comes out as \(3.08 \times 10^{15} \text{s}^{-1}\), indicating significant electromagnetic radiation.

This frequency calculation is crucial in linking physical phenomena such as electron transitions with observable properties of light like color or wavelength.

Ionization Energy

Ionization energy is the energy required to move an electron from the bound state in an atom to a free state, effectively ionizing the atom. In the context of the hydrogen atom, it is the energy needed to take an electron completely away from the nucleus.Given in our exercise is an ionization energy value of \(2.18 \times 10^{-18} \text{ J atom}^{-1}\).

• This value represents the energy for an electron to go from the ground state (\(n=1\)) to \(n=\infty\) or a free electron state.
• The ionization energy provides a reference point when discussing electron transitions and energy calculations.

Understanding ionization energy helps us appreciate the effort needed to liberate an electron and underpins several atomic and quantum behavior principles. It also forms the basis of analyzing different elements by comparing their unique ionization potentials.