Which is that famous helium ion problem in IIT-JEE 1992(or 1993)?

Can anyone post it ?

It rated as being one of the toughest problems.

The famous helium ion problem from the IIT-JEE exam in 1992 is indeed one of the most challenging questions that has been discussed among students and educators alike. This problem revolves around the quantum mechanics of the helium ion (He⁺), which is a hydrogen-like atom with two protons and one electron. The complexity arises from the need to consider both the electrostatic interactions and the quantum mechanical principles governing the behavior of electrons in multi-electron systems.

Understanding the Helium Ion Problem

The specific problem typically involves calculating the energy levels of the helium ion or determining the wavelength of emitted radiation during electronic transitions. The difficulty lies in the fact that while the helium ion has only one electron, the presence of two protons creates a more complicated potential field than that of hydrogen.

Key Concepts to Consider

• Quantum Mechanics: The behavior of electrons in atoms is described by quantum mechanics, which introduces concepts such as wave functions and energy quantization.
• Electrostatic Forces: The interaction between the positively charged nucleus and the negatively charged electron must be considered, as well as the repulsion between multiple protons.
• Bohr Model Adaptation: While the Bohr model is primarily used for hydrogen, it can be adapted for hydrogen-like ions by modifying the effective nuclear charge.

Solving the Problem

To tackle the helium ion problem, one typically follows these steps:

1. Identify the Effective Nuclear Charge: For He⁺, the effective nuclear charge (Z) is 2, since there are two protons in the nucleus.
2. Apply the Bohr Model Formula: The energy levels for a hydrogen-like atom can be calculated using the formula:
   E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}
   where n is the principal quantum number.
3. Calculate Energy Levels: For example, for n=1, the energy would be:
   E_1 = -\frac{2^2 \cdot 13.6}{1^2} = -54.4 \text{ eV}
4. Determine Wavelength of Emission: If an electron transitions from a higher energy level to a lower one, the energy difference can be used to find the wavelength of emitted radiation using the formula:
   \(\Delta E = h \cdot f\) and \(c = \lambda \cdot f\), where h is Planck's constant, c is the speed of light, and λ is the wavelength.

Example Calculation

Suppose an electron transitions from n=2 to n=1. The energy levels would be:

• E_2 = -\frac{2^2 \cdot 13.6}{2^2} = -13.6 \text{ eV}
• E_1 = -54.4 \text{ eV}

The energy difference (ΔE) is:

ΔE = E_1 - E_2 = -54.4 - (-13.6) = -40.8 eV

Using the energy-wavelength relationship, we can find the wavelength of the emitted photon:

λ = \frac{hc}{ΔE}

By substituting the values for h and c, you can calculate the wavelength.

Final Thoughts

This problem not only tests your understanding of quantum mechanics and atomic structure but also your ability to apply theoretical concepts to practical calculations. It’s a classic example of how complex interactions in atomic systems can lead to intricate problems that challenge even the most prepared students. Mastering such problems requires a solid grasp of the underlying principles and practice with similar questions.