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Chemistry-Ionization Energy

NEET Chemistry Ionization Energy MCQ Question

Type: MCQ-numerical-Hard-Class 11

Calculate the ionization enthalpy of atomic hydrogen in kJ mol⁻¹ using the energy of its electron in the ground state given as –2.18×10⁻¹⁸ J. (1 J = 10⁻³ kJ, Avogadro's number = 6.022×10²³ mol⁻¹)

A

1312

B

2180

C

1090

D

436

Correct Answer

Option A

Detailed Explanation

The ionization enthalpy can be calculated by multiplying the energy of the electron (-2.18×10⁻¹⁸ J) by Avogadro's number (6.022×10²³ mol⁻¹) and converting J to kJ. This results in approximately 1312 kJ mol⁻¹.

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