In Bohr's atomic model, the energy of the ele | vedclass

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

Question

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

Vedclass · Accepted Answer

$16E/9$

Vedclass · Answer

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For the second orbit of a hydrogen atom $(Z=1, n=2)$: $E = -13.6 	imes \frac{1^2}{2^2} = -13.6 	imes \frac{1}{4} 	ext{ eV}$.Thus, $-13.6 	ext{ eV} = 4E$.For the third orbit of a helium atom $(Z=2, n=3)$: $E' = -13.6 	imes \frac{2^2}{3^2} = -13.6 	imes \frac{4}{9} 	ext{ eV}$.Substituting $-13.6 = 4E$ into the equation for $E'$:$E' = (4E) 	imes \frac{4}{9} = \frac{16E}{9}$.

In Bohr's atomic model, the energy of the electron is $E$ in the second orbit of a hydrogen atom. The energy of the electron in the third orbit of a helium atom $(Z = 2)$ will be:

Similar Questions

The angular momentum of an electron in the $2^{nd}$ excited state of a Helium ion $(He^+)$ is

The ratio of wavelengths for the transition of electrons from the $2^{nd}$ orbit to the $1^{st}$ orbit of Helium $(He^+)$ and Lithium $(Li^{++})$ is (Atomic number of Helium = $2$,Atomic number of Lithium = $3$).

If the first excitation potential of a hydrogen-like atom is $V \, \text{volt}$, then the ionization energy of this atom will be

The de-Broglie wavelength of an electron in the first Bohr orbit is

The ratio of the ionization energy of a Bohr's hydrogen atom to that of a Bohr's hydrogen-like lithium atom is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module