Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \, J$. Calculate the ionization enthalpy of atomic hydrogen in terms of $J \, mol^{-1}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
The energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \, J$.
The energy required to remove this electron to infinity (ionization) is the negative of the ground state energy,which is $2.18 \times 10^{-18} \, J$ per atom.
To calculate the ionization enthalpy in $J \, mol^{-1}$,we multiply the energy per atom by the Avogadro constant $(N_A = 6.022 \times 10^{23} \, mol^{-1})$:
$\text{Ionization enthalpy} = 2.18 \times 10^{-18} \, J \times 6.022 \times 10^{23} \, mol^{-1} = 1.312 \times 10^{6} \, J \, mol^{-1}$.

Explore More

Similar Questions

Which one of the following frequencies of radiation (in $Hz$) has a wavelength of $600 \ nm$?

The process of the isolation of a metal by dissolving the ore in a suitable chemical reagent followed by precipitation of the metal by a more electropositive metal is called:

If $f: N \rightarrow Z$ is defined by $f(n)=\begin{cases} 2 & \text{if } n=3k, k \in Z \\ 10 & \text{if } n=3k+1, k \in Z \\ 0 & \text{if } n=3k+2, k \in Z \end{cases}$,then $\{n \in N: f(n)>2\}$ is equal to

Which artery is absent in a frog?

$N_2$ and $O_2$ are converted into monocations $N_2^+$ and $O_2^+$ respectively. Which statement is wrong?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo