In Bohr's model of hydrogen-like species,whic | vedclass

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

Question

(A) The kinetic energy $(KE)$ of an electron in the $n$-th orbit of a hydrogen-like species with atomic number $Z$ is given by $KE = 13.6 	imes \frac

Vedclass · Accepted Answer

$KE$ of electron in ground state of hydrogen atom is equal to $KE$ of electron in $n = 3$ state of $Li^{2+}$

Vedclass · Answer

(A) The kinetic energy $(KE)$ of an electron in the $n$-th orbit of a hydrogen-like species with atomic number $Z$ is given by $KE = 13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For hydrogen atom $(Z=1)$,$KE$ in ground state $(n=1)$ is $13.6 	imes \frac{1^2}{1^2} = 13.6 	ext{ eV}$.For $Li^{2+}$ $(Z=3)$,$KE$ in $n=3$ state is $13.6 	imes \frac{3^2}{3^2} = 13.6 	ext{ eV}$.Comparing these,the $KE$ of an electron in the ground state of a hydrogen atom is equal to the $KE$ of an electron in the $n=3$ state of $Li^{2+}$.Thus,option $A$ is correct.

In Bohr's model of hydrogen-like species,which of the following is true?

Similar Questions

In a hydrogen atom,an electron makes a transition from the $n=4$ to the $n=1$ energy level. The recoil momentum of the $H$ atom will be:

The electron in a hydrogen atom makes a transition from the $M$ shell to the $L$ shell. The ratio of the magnitudes of the initial to the final centripetal acceleration of the electron is:

The electron in a hydrogen atom is moving in an orbit of radius $0.53 \text{ Å}$. It takes $1.571 \times 10^{-16} \text{ s}$ to complete one revolution. The velocity of the electron will be $[\pi = 3.142]$.

The ratio between the total acceleration of the electron in a singly ionized helium atom and a hydrogen atom (both in the ground state) is:

In Bohr's model,the atomic radius of the first orbit is $r_0$,then the radius of the third orbit is

Vedclass Test Series

Exam Paper Generator

Online Exam Module