The ionisation energy of a hydrogen atom is 13.6 eV. Using 1 eV = 1.6x10^-19 J and N_A = 6.022x10^23 mol^-1, express it in kJ mol^-1 (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3as081

The ionisation energy of a hydrogen atom is $13.6 eV$ . Using $1 eV = 1.6 \times 1 0^{- 19} J$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , express it in $kJ mo l^{- 1}$ (nearest integer).

Numerical answer type, answer in kJ/mol, tolerance 2.

Energy per atom

= 13.6 \times 1.6 \times 1 0^{- 19} = 2.176 \times 1 0^{- 18} J

. Multiplying by

N_{A}

gives

1310.4 kJ mo l^{- 1}

, i.e. 1310.