Ionisation energy of He^{+} is 19.6 times 10^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Ionisation energy $(IE)$ is the energy required to move an electron from the ground state to infinity.$IE = E_{\infty} - E_{1} = 0 - E_{1} = -E_{1

Vedclass · Accepted Answer

$-4.41 	imes 10^{-17} \, J \, atom^{-1}$

Vedclass · Answer

(B) Ionisation energy $(IE)$ is the energy required to move an electron from the ground state to infinity.$IE = E_{\infty} - E_{1} = 0 - E_{1} = -E_{1}$Therefore,$E_{1}$ of $He^{+} = -19.6 	imes 10^{-18} \, J \, atom^{-1}$.Energy of a hydrogen-like species at state $n$ is given by $(E_{n})_{species} = (E_{1})_{H} 	imes \frac{Z^{2}}{n^{2}}$.For $He^{+}$ $(Z=2, n=1)$: $(E_{1})_{He^{+}} = (E_{1})_{H} 	imes 2^{2} = -19.6 	imes 10^{-18} \, J \, atom^{-1}$.So,$(E_{1})_{H} = \frac{-19.6 	imes 10^{-18}}{4} \, J \, atom^{-1}$.For $Li^{2+}$ $(Z=3, n=1)$: $(E_{1})_{Li^{2+}} = (E_{1})_{H} 	imes 3^{2} = \frac{-19.6 	imes 10^{-18}}{4} 	imes 9 = -4.41 	imes 10^{-17} \, J \, atom^{-1}$.

Ionisation energy of $He^{+}$ is $19.6 \times 10^{-18} \, J \, atom^{-1}$. The energy of the first stationary state $(n = 1)$ of $Li^{2+}$ is

Similar Questions

Find the wave number of the shortest wavelength in the Balmer series.

Calculate the energy in joule corresponding to light of wavelength $45 \ nm$. (Planck's constant,$h = 6.63 \times 10^{-34} \ J \ s$,speed of light,$c = 3 \times 10^8 \ m \ s^{-1}$)

Explain the hydrogen spectrum using Bohr's model.

In an emission spectrum,the transitions $n_2 \to n_1$,$n_3 \to n_2$,$n_4 \to n_3$,and $n_5 \to n_4$ emit energies $E_1, E_2, E_3$,and $E_4$ respectively. What is the increasing order of these energies?

The radius of the fourth orbit in $He^{+}$ ion is '$R_1$' $pm$ and the radius of the third orbit in $Li^{2+}$ ion is '$R_2$' $pm$. The value of $(R_1 - R_2)$ in $pm$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module