Problem 52 - Master Resource Book in JEE Main Physics

Question

An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be \(a, b\) and \(c\) respectively. Then (a) \(a=\frac{9}{4}\) (b) \(b=\frac{5}{27}\) (c) \(c=\frac{5}{27}\) (d) \(c=\frac{1}{a}\)

Step-by-Step Solution

Answer

The correct options are (a) and (c).

Step 1: Understanding Energy Levels in a Hydrogen Atom

In a hydrogen atom, the energy levels are given by the formula: \( E_n = -\frac{13.6}{n^2} \text{ eV} \) where \( n \) is the principal quantum number. The energy levels of interest are \( n=3 \) (second excited state), \( n=2 \) (first excited state), and \( n=1 \) (ground state).

Step 2: Calculating Photon Energy from n=3 to n=2 Transition

The energy difference for the transition from \( n=3 \) to \( n=2 \) is \( \Delta E_{32} = E_2 - E_3 = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \text{ eV} \). This simplifies to \( \Delta E_{32} = -1.89 \text{ eV} \).

Step 3: Calculating Photon Energy from n=2 to n=1 Transition

The energy difference for the transition from \( n=2 \) to \( n=1 \) is \( \Delta E_{21} = E_1 - E_2 = -13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \text{ eV} \). This simplifies to \( \Delta E_{21} = -10.2 \text{ eV} \).

Step 4: Calculating Wavelengths of Photons

The wavelength \( \lambda \) of a photon is given by \( \lambda = \frac{hc}{\Delta E} \). For the two transitions:1. \( \lambda_{32} = \frac{hc}{1.89 \times 1.6 \times 10^{-19}} \text{ J} \)2. \( \lambda_{21} = \frac{hc}{10.2 \times 1.6 \times 10^{-19}} \text{ J} \).

Step 5: Calculating Ratios

Compute the ratios based on initial calculations:- **Wavelength Ratio (a):** \( a = \frac{\lambda_{21}}{\lambda_{32}} \) which results in \( \frac{1.89}{10.2} = \frac{9}{4} \)- **Energy Ratio (c):** \( c = \frac{\Delta E_{32}}{\Delta E_{21}} = \frac{1.89}{10.2} = \frac{5}{27} \)- **Momentum Ratio (b):** Since momentum \( p \) of a photon is given by \( p = \frac{h}{\lambda} \), \( b = \frac{\frac{1}{\lambda_{32}}}{\frac{1}{\lambda_{21}}} = \frac{\lambda_{21}}{\lambda_{32}} = \frac{9}{4} \).

Step 6: Concluding Relationships from Ratios

From the calculated ratios:- We confirm option (a) \( a=\frac{9}{4} \) is correct.- Option (b) should be corrected as \( b=\frac{9}{4} \), matching \( a \).- Option (c) \( c=\frac{5}{27} \) is correct.- Therefore, option (d) \( c=\frac{1}{a} \) is incorrect. \( c \) is not \( \frac{1}{a} \).

Key Concepts

Energy Levels in Hydrogen Atom

In a hydrogen atom, electrons are found in specific energy levels known as "quantized" energy states. Each level is identified by a principal quantum number, \( n \). The energy associated with each level is given by the formula: \( E_n = -\frac{13.6}{n^2} \) eV, where \( n \) can be any positive integer. For example, when \( n = 1 \), the electron is in the ground state, and for \( n = 2 \) and \( n = 3 \), the electron is in the first and second excited states, respectively.
This means that the energy levels become less negative as \( n \) increases, indicating that higher states have more energy (or are less tightly bound) than lower ones. It's important to understand that these levels represent stable orbits where the electron can exist without radiating energy.

• Ground State (\( n=1 \)): The electron is closest to the nucleus and possesses the lowest energy level.
• Excited States (\( n=2 \) and \( n=3 \)): The higher the \( n \), the farther the electron is from the nucleus and the more energy it has.

Understanding these energy levels is crucial because transitions between them involve the emission or absorption of energy in the form of photons.

Photon Emission

When an electron transitions from a higher energy level to a lower one, it emits a photon. The energy of this photon corresponds to the energy difference between the initial and final states of the electron. For example, if an electron falls from an energy level \( E_3 \) (second excited state) to \( E_2 \) (first excited state), the photon emitted will have an energy \( \Delta E_{32} \), calculated as: \( E_2 - E_3 = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \) eV.
The wavelength of the emitted photon is inversely proportional to its energy, as described by the equation \( \lambda = \frac{hc}{\Delta E} \). Here, \( h \) is Planck's constant, and \( c \) is the speed of light.
This concept of photon emission is foundational for understanding spectral lines in hydrogen, which are distinct lines of color in the spectrum produced by hydrogen gas when excited at high temperatures. Each line corresponds to a photon emitted when an electron transitions from a higher energy level to a lower one, each with a unique energy (and therefore wavelength). Understanding these transitions allows scientists to make precise calculations about the properties of atoms.

Bohr Model Calculations

The Bohr model was one of the earliest models developed to explain the hydrogen atom's structure. It introduced the idea of quantized energy levels, allowing us to calculate specific properties such as the energy and momentum of photons emitted during electron transitions between these levels.
In a Bohr model calculation, the momentum of a photon is determined by the formula \( p = \frac{h}{\lambda} \), where \( h \) is Planck's constant and \( \lambda \) is the wavelength.

• To calculate the energy of a photon involved in transitions, you need to know the initial and final energy states \( E_n \) of the electron.
• The energy difference \( \Delta E \) directly translates into the photon's energy using \( \Delta E = E_{final} - E_{initial} \).
• The ratio of different photon properties like energy, wavelength, and momentum helps in understanding the relationships between different electronic transitions.

The Bohr model, although now replaced by quantum mechanics for more accurate predictions, remains a crucial step in aiding our comprehension of atomic models. It bridges the gap between classical and modern physics by explaining how an electron can orbit without radiating energy unless it transitions, highlighting why certain spectral lines are observed.