Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10^–18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^–1.

Open in App

Solution

It is given that the energy of an electron in the ground state of the hydrogen atom is –2.18 × 10^–18J.

Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 × 10^–18J.

Ionization enthalpy of atomic hydrogen = 2.18 × 10^–18J

Hence, ionization enthalpy of atomic hydrogen in terms of J mol^–1 = 2.18 × 10^–18 × 6.02 × 10²³ J mol^–1 = 1.31 × 10⁶ J mol^–1

Suggest Corrections

Similar questions

Q. Energy of an electron in the ground state of the hydrogen atom is -

2.18 \times 10^{- 18} J .

The ionization enthalpy of atomic hydrogen is

a \times 10^{5} J / m o l

. Then

‘ a^{'}

is: (Take

N_{A} = 6 \times 10^{23}

)

Q. The ionization energy of the ground state of hydrogen atom is

2.18 \times 10^{- 18}

J. The energy of an electron in its second orbit would be:

Calculate the energy required for the process
$H e_{(g)}^{+} \to H e_{(g)}^{2 +} + e^{-}$
The ionization energy for the H atom in the ground state is 2.18 × $10^{- 18} J a t o m^{- 1}$

Calculate the energy required for the process

The ionization energy for the H atom in the ground state is 2.18 ×10^–18 J atom^–1

The energy of an electron in the ground state of a $H$ -atom is $- 2.15 \times 10^{- 18} J$ . The Ionisation energy of the atomic hydrogen in terms of $10^{5} J m o l^{- 1}$ is:

Join BYJU'S Learning Program

Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1.

Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10^–18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^–1.