If the ionization energy of He^{+} is 8.68 ti | vedclass

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

Question

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$ or $

Vedclass · Accepted Answer

$-8.68 	imes 10^{-18} \ J$

Vedclass · Answer

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$ or $E_n = -K 	imes \frac{Z^2}{n^2} \ J$. For $He^{+}$ $(Z=2)$,the ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$. $IE = E_{\infty} - E_1 = 0 - (-K 	imes \frac{2^2}{1^2}) = 4K = 8.68 	imes 10^{-18} \ J$. Therefore,$K = \frac{8.68 	imes 10^{-18}}{4} = 2.17 	imes 10^{-18} \ J$. Now,for $Be^{3+}$ $(Z=4)$ in the second orbit $(n=2)$: $E_2 = -K 	imes \frac{Z^2}{n^2} = -(2.17 	imes 10^{-18}) 	imes \frac{4^2}{2^2} = -(2.17 	imes 10^{-18}) 	imes \frac{16}{4} = -(2.17 	imes 10^{-18}) 	imes 4 = -8.68 	imes 10^{-18} \ J$.

If the ionization energy of $He^{+}$ is $8.68 \times 10^{-18} \ J$,then the energy of $Be^{3+}$ ion in the second orbit is:- ($Z$ of $Be = 4$)

Similar Questions

What is the maximum number of emission lines when the excited electron of an $H$ atom in $n=6$ drops to the ground state?

What is the relation between radius, order of orbit, and nuclear charge for hydrogen-like species?

If the longest wavelength of the spectral line of the Paschen series of $Li^{2+}$ ion spectrum is $x \ \mathring{A}$,then the longest wavelength (in $\mathring{A}$) of the Lyman series of the hydrogen spectrum is

Which of the following orders is $NOT$ correct for the $1^{st}$,$2^{nd}$,and $3^{rd}$ orbits of a hydrogen-like atom?

When light is directed at the metal surface,the emitted electrons

Vedclass Test Series

Exam Paper Generator

Online Exam Module