The energy of an electron in the first orbit | vedclass

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

Question

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 	imes 10^{-18} \ J 	imes \frac{

Vedclass · Accepted Answer

$-217.9 	imes 10^{-20} \ J$

Vedclass · Answer

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 	imes 10^{-18} \ J 	imes \frac{Z^2}{n^2}$.For $He^+$,$Z = 2$ and $n = 1$,so $E_{1, He^+} = E_{1, H} 	imes Z^2$.Given $E_{1, He^+} = -871.6 	imes 10^{-20} \ J$.Therefore,$-871.6 	imes 10^{-20} = E_{1, H} 	imes (2)^2$.$-871.6 	imes 10^{-20} = E_{1, H} 	imes 4$.$E_{1, H} = \frac{-871.6 	imes 10^{-20}}{4} = -217.9 	imes 10^{-20} \ J$.

The energy of an electron in the first orbit of $He^+$ is $-871.6 \times 10^{-20} \ J$. The energy of an electron in the first orbit of hydrogen is ...... .

Similar Questions

The energy levels for a hydrogen-like species with atomic number $Z$ can be given by:

If $n=2$ for $He^{+}$ ion,then what is the wavelength in $\mathring{A}$?

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

With increasing principal quantum number,the energy difference between adjacent energy levels in $H$ atom

The limiting line in the Balmer series will have a frequency of (Rydberg constant,$R_{\infty} = 3.29 \times 10^{15} \ s^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module