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 \(\left.\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}\right)\) (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 radiation emitted is \(3.08 \times 10^{15} \, \text{s}^{-1}\) (option c).

Step 1: Calculate Energy Levels

We use the formula for the energy levels of a hydrogen atom: \(E_n = -\frac{2.18 \times 10^{-18} \, \text{J}}{n^2} \). Calculate the energies for \(n=4\) and \(n=1\).For \(n=4\), \(E_4 = -\frac{2.18 \times 10^{-18} \, \text{J}}{4^2} = -1.3625 \times 10^{-19} \, \text{J}\).For \(n=1\), \(E_1 = -\frac{2.18 \times 10^{-18} \, \text{J}}{1^2} = -2.18 \times 10^{-18} \, \text{J}\).

Step 2: Calculate the Energy Change

Find the energy change \(\Delta E\) when the electron falls from \(n=4\) to \(n=1\): \(\Delta E = E_1 - E_4 = (-2.18 \times 10^{-18} \, \text{J}) - (-1.3625 \times 10^{-19} \, \text{J}) = -2.04375 \times 10^{-18} \, \text{J}\).

Step 3: Calculate Frequency from Energy Change

Using Planck's equation \(\Delta E = h u \), solve for the frequency \(u\):\[ u = \frac{\Delta E}{h} = \frac{-2.04375 \times 10^{-18} \, \text{J}}{6.625 \times 10^{-34} \, \text{Js}} \approx 3.08 \times 10^{15} \, \text{s}^{-1} \].

Step 4: Identify the Correct Answer

The calculated frequency, \(3.08 \times 10^{15} \, \text{s}^{-1}\), matches option (c). Thus, the frequency of radiation emitted when the electron falls from \(n=4\) to \(n=1\) is \(3.08 \times 10^{15} \, \text{s}^{-1}\).

Key Concepts

Energy Level Calculation

In a hydrogen atom, electrons reside in specific energy levels, or orbits, around the nucleus. Each of these energy levels can be determined using the equation:\[E_n = -\frac{2.18 \times 10^{-18} \, \text{J}}{n^2}\]Where:

• \(E_n\) is the energy of the electron at level \(n\)
• \(2.18 \times 10^{-18} \, \text{J}\) is the ionization energy of hydrogen
• \(n\) is the principal quantum number

To find the energies for specific levels, such as when \(n=4\) and \(n=1\), simply plug the values into the formula:

• For \(n=4\), \(E_4 = -\frac{2.18 \times 10^{-18} \, \text{J}}{4^2} = -1.3625 \times 10^{-19} \, \text{J}\)
• For \(n=1\), \(E_1 = -\frac{2.18 \times 10^{-18} \, \text{J}}{1^2} = -2.18 \times 10^{-18} \, \text{J}\)

This calculation is crucial as it helps determine the energy change when electrons transit between levels.

Frequency Calculation

The frequency of radiation emitted by an electron transitioning between energy levels can be found by first calculating the energy difference. This change, known as \(\Delta E\), is crucial because it represents the energy released or absorbed as the electron moves. Calculate \(\Delta E\) using the energy levels identified:\[\Delta E = E_1 - E_4 = (-2.18 \times 10^{-18} \, \text{J}) - (-1.3625 \times 10^{-19} \, \text{J}) = -2.04375 \times 10^{-18} \, \text{J}\]With \(\Delta E\) known, we can use it to determine the frequency, \(u\), of the emitted radiation.The magnitude of \(\Delta E\) tells us how much energy is "used" to produce a wave with a certain frequency. This frequency is obtained using Planck's equation, making it directly proportional to the energy change.

Planck's Equation

Planck's equation connects the change in energy \((\Delta E)\) during an electron transition with the frequency of the radiation produced. The equation is given by:\[\Delta E = h \cdot u\]Where:

• \(\Delta E\) is the energy change from the transition
• \(h = 6.625 \times 10^{-34} \, \text{Js}\) is Planck's constant
• \(u\) is the frequency of the emitted radiation

Planck's equation tells us that the energy change \(\Delta E\) equals the product of Planck's constant \(h\) and the frequency \(u\). To find the frequency \(u\), rearrange the equation:\[u = \frac{\Delta E}{h}\]Substituting the values:\[ u = \frac{-2.04375 \times 10^{-18} \, \text{J}}{6.625 \times 10^{-34} \, \text{Js}} \approx 3.08 \times 10^{15} \, \text{s}^{-1} \]This enables us to identify the frequency of the radiation that results from the transition, which in this context is a crucial component in understanding atomic spectra.