Verified Solution • 2025–26 Session moderate

QJEEM08Ch46dd: JEE Main 2008 Chemistry - Class 11 JEEM08Ch46dd Solution: The ionization enthalpy of hydrogen atom is 1.312 × 10 6 J mol…

The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is

Multiple Choice Options

  • A 8.51 × 105 J mol−1
  • B 6.56 × 105 J mol−1
  • C 7.56 × 105 J mol−1
  • D 9.84 × 105 J mol−1

Step-by-Step Verified Solution

Note :

1 eV/atom = 96.485 ×\times 103 J/mol

∴\therefore 13.6 eV/atom = 13.6 ×\times 96.485 ×\times 103 J/mol = 1.312 × 106 J mol−1

Energy required to excite the electron from n1 to n2 is

ΔE=13.6×Z2(1n12−1n22)\Delta E = 13.6 \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

= 1.312 × 106 × 1 (112−122)\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)

= 1.312 × 34{3 \over 4} × 106

= 9.84 × 105 J mol−1

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