The energy and radius of an electron present | vedclass

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

Question

(D) For $He^{+}$ ion, the atomic number $Z = 2$ and the orbit number $n = 2$. The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18

Vedclass · Accepted Answer

$-2.18 	imes 10^{-18} \ J, 105.8 \ pm$

Vedclass · Answer

(D) For $He^{+}$ ion, the atomic number $Z = 2$ and the orbit number $n = 2$. The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}{n^2} \ J$. Substituting the values: $E_2 = -2.18 	imes 10^{-18} 	imes \frac{2^2}{2^2} = -2.18 	imes 10^{-18} \ J$. The radius of the $n^{th}$ orbit is given by $r_n = 52.9 	imes \frac{n^2}{Z} \ pm$. Substituting the values: $r_2 = 52.9 	imes \frac{2^2}{2} = 52.9 	imes 2 = 105.8 \ pm$. Thus, the energy is $-2.18 	imes 10^{-18} \ J$ and the radius is $105.8 \ pm$.

The energy and radius of an electron present in the second orbit of $He^{+}$ are respectively:

Similar Questions

The total energy of the electron in the hydrogen atom in the ground state is $-13.6 \, eV$. What is the stopping potential in volts required to remove this electron from the ground state?

The shortest wavelength in the hydrogen spectrum is approximately (in $nm$)

What is the difference in the line spectrum of different atoms compared to the hydrogen atom?

For the hydrogen emission spectrum,the energy change $E$ (in Joules) is given by $\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ J$,where $n_1 = 1, 2, 3, \dots$ and $n_2 = 2, 3, 4, \dots$. Which of the following corresponds to the Paschen series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module