Problem 39 - Master Resource Book in JEE Main Physics

Question

An electron jumps from the 4 th orbit to 2 nd orbit of hydrogen atom. Given the Rydberg's constant \(R=10^{5} \mathrm{~cm}^{-1}\), the frequency in hertz of the emitted radiation will be (a) \(\frac{3}{16} \times 10^{5}\) (b) \(\frac{3}{16} \times 10^{15}\) (c) \(\frac{9}{16} \times 10^{15}\) (d) \(\frac{3}{4} \times 10^{15}\)

Step-by-Step Solution

Answer

The frequency of the emitted radiation is \( \frac{9}{16} \times 10^{15} \) Hz (option c).

Step 1: Understand the Concept

When an electron transitions from a higher orbit to a lower orbit within a hydrogen atom, it emits radiation. The frequency of this emitted radiation can be determined using the Rydberg formula for hydrogen.

Step 2: Recall the Rydberg Formula

The Rydberg formula is given by:\[u = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]where \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively. In this case, \( n_1 = 2 \) and \( n_2 = 4 \).

Step 3: Substitute Values into the Rydberg Formula

Substitute the given values into the Rydberg formula:\[u = 10^{5} \left( \frac{1}{2^2} - \frac{1}{4^2} \right)\]Calculate the values inside the parentheses.

Step 4: Calculate the Transition Energy

First, calculate the individual fractions:\[\frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{4^2} = \frac{1}{16}\]Subtract to find the difference:\[\frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16}\]

Step 5: Calculate the Frequency

Now, substitute this result back into the formula to find the frequency:\[u = 10^{5} \times \frac{3}{16} = \frac{3}{16} \times 10^{5} \text{ cm}^{-1}\]To convert to Hz, we multiply by the speed of light \(c\) (\(3 \times 10^{10}\, \text{cm/s}\)):\[u = \left(\frac{3}{16} \times 10^{5}\right) \times (3 \times 10^{10}) = \frac{3}{16} \times 3 \times 10^{15}\]

Step 6: Simplify the Frequency Expression

Simplify the expression calculated:\[u = \frac{3 \times 3}{16} \times 10^{15} = \frac{9}{16} \times 10^{15}\]

Key Concepts

Electron Transition

When talking about electron transitions, we are diving into a fascinating aspect of atomic physics. In simple terms, an electron transition refers to the movement of an electron from one energy level or orbit to another within an atom. These energy levels are often visualized as orbits around the nucleus.

For a hydrogen atom, these orbits are defined by principal quantum numbers such as 1, 2, 3, and so on. When an electron "jumps" from a higher energy level (like the 4th) to a lower one (like the 2nd), energy is emitted in the form of radiation. This emitted energy corresponds to a specific wavelength or frequency of light. Understanding this process is vital because it explains how light is produced on an atomic scale and ties closely with the Rydberg formula, which uses this principle to predict the wavelengths of spectral lines.

Hydrogen Atom Spectrum

The hydrogen atom spectrum is a great example to learn from when studying light emission caused by electron transitions. The most well-known series in the hydrogen spectrum include the Lyman, Balmer, and Paschen series. These are sets of wavelengths emitted by hydrogen atoms, and each series corresponds to transitions ending at a specific energy level.

- In the Lyman series, all transitions end at the n=1 level, emitting ultraviolet light. - The Balmer series involves transitions ending at n=2, resulting in visible light emission. - The Paschen series involves transitions to the n=3 level, emitting infrared light.
The spectrum of hydrogen is particularly simple and distinct because hydrogen has only one electron, which transitions between energy levels creating these spectral lines. Observing this spectrum in a laboratory setting or a simulation can provide a visual assistance to better grasp how electrons shift energy levels.

Frequency Calculation

The calculation of frequency in the context of electron transitions uses the Rydberg formula. This formula is a powerful tool that relates the frequency of light emitted (or absorbed) as an electron makes a transition between energy levels. The core of frequency calculation using this formula revolves around the following:
\[u = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]
Here, \( u \) is the frequency, \(R_H\) is the Rydberg constant (approximately \(10^5 \, \mathrm{cm}^{-1}\) for this context), and \(n_1\) and \(n_2\) are the principal quantum numbers of the electron's initial and final energy levels, respectively.

Calculating the difference \(\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\) provides the change in energy levels, which when multiplied by the Rydberg constant gives the wave number. This is then converted to frequency in hertz by multiplying with the speed of light. This entire process explains why transitions between different energy levels in hydrogen produce specific frequencies of light, hence causing the distinct spectral lines observed.