The ionization energy of a singly ionized hel | vedclass

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

Question

(A) For a hydrogen-like species,the total energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $He^+

Vedclass · Accepted Answer

Kinetic energy of the first orbit

Vedclass · Answer

(A) For a hydrogen-like species,the total energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $He^+$,$Z = 2$ and for the ground state,$n = 1$. Thus,$E_1 = -13.6 	imes \frac{2^2}{1^2} = -54.4 \ eV$. The ionization energy $(IE)$ is the energy required to remove the electron,which is $-E_1 = 54.4 \ eV$. According to Bohr's model,the kinetic energy $(KE)$ of an electron in the $n^{th}$ orbit is equal to the magnitude of its total energy $(|E_n|)$. Therefore,$IE = |E_1| = KE_1$.

The ionization energy of a singly ionized helium ion $(He^+)$ is equal to:

Similar Questions

The emission spectrum of hydrogen discovered first and the region of the electromagnetic spectrum in which it belongs, respectively, are

The ionization energy of a hydrogen atom is $13.6 \ eV$. The energy required to ionize a hydrogen atom in its second excited state is ........... $eV$.

Find out the distance between the $2^{nd}$ and $3^{rd}$ orbit of the $H^{-}$ ion.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [$a_0$ is Bohr radius]:

Bohr’s model of atom explains :

Vedclass Test Series

Exam Paper Generator

Online Exam Module