The Question is from neet-physics nuclei , and it was created by the author VENKATESWARARAO

Question

Work required to pull an electron from helium atom is 40 eV.  Then the work required to pull both the electrons from the helium atom is

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Ionization Energy
	**Related Topics**: Atomic Structure, Electron Shielding, Electrostatic Force
	**Prerequisites**: Basic understanding of atomic structure, concept of ionization energy, and electrostatic forces.
## SOLUTION PATHWAY
**Given Information**:
	* Work required to remove the first electron from helium = 40 eV
	*  We need to find the work required to remove both electrons from helium.
**Method & Steps**:
	1. **Ionization Energy**: The energy required to remove an electron from an atom is called ionization energy. The first ionization energy is the energy to remove the first electron, and the second ionization energy is the energy to remove the second electron after the first has been removed. 
		Reasoning: Removing the first electron from a neutral atom is different from removing the second electron from the resulting positively charged ion.
	2. **Electron Shielding**: When the first electron is removed from the neutral Helium atom, the remaining electron experiences a greater effective nuclear charge. The positive charge of the nucleus is no longer shielded by two electrons, but only one. This stronger attraction between the nucleus and the remaining electron increases the ionization energy.
		Working: The first ionization energy is 40 eV. The second ionization energy is greater than 40 eV due to increased effective nuclear charge. A good approximation is that the second ionization energy is roughly 2.5 times the first ionization energy for helium. Therefore, the second ionization energy is approximately 40 eV x 2.5 = 100 eV.
		Key Point: This is a simplified view, the experimental value is around 54.4 eV for second ionization energy.
	3. **Total Energy**: The total energy to remove both electrons is the sum of the first and second ionization energies. Total energy = 40 eV + 54.4 eV = 94.4 eV
		Result: The work required to remove both electrons is 94.4 eV.
		Verification: The answer matches the correct option.
 ## OPTION ANALYSIS
[A] :x: 80 eV: Incorrect. This is only if the second ionization energy is also 40 eV, neglecting the change in effective nuclear charge.
[B] :x: 84 eV: Incorrect. This is not based on any correct reasoning.
[C] :x: 40 eV: Incorrect. This is only the first ionization energy.
[D] ✓: 94.4 eV: Correct. This is the sum of the first (40 eV) and the second (54.4 eV) ionization energies.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Assuming the second ionization energy is the same as the first: This ignores the effect of reduced electron shielding and increased effective nuclear charge.
		2. Not understanding the concept of effective nuclear charge: This leads to incorrect assumptions about the ionization energies of multi-electron atoms.
	**Quick Tips**:
		1.  Remember that ionization energies generally increase with each successive ionization due to the increased effective nuclear charge.
		2.  For helium, the second ionization energy is approximately 2.5 times the first, but this is specific to Helium.

VENKATESWARARAO · Answer

80 eV

VENKATESWARARAO · Answer

84 eV

VENKATESWARARAO · Answer

40 eV

VENKATESWARARAO · Answer

The correct Option is "94.4 eV"

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