The ratio of the speed of the electron in the | vedclass

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

Question

(D) The speed of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula $v_n = v_0 \frac{Z}{n}$,where $Z$ is the atomic num

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(D) The speed of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula $v_n = v_0 \frac{Z}{n}$,where $Z$ is the atomic number and $n$ is the principal quantum number.Since the principal quantum number $n$ is the same $(n=3)$ for both cases,the speed is directly proportional to the atomic number $Z$ $(v \propto Z)$.For $He^{+}$,the atomic number $Z_{He^+} = 2$.For the hydrogen atom $(H)$,the atomic number $Z_H = 1$.Therefore,the ratio of the speeds is $\frac{v_{He^+}}{v_H} = \frac{Z_{He^+}}{Z_H} = \frac{2}{1} = 2:1$.

The ratio of the speed of the electron in the $3^{rd}$ orbit of $He^{+}$ to the speed of the electron in the $3^{rd}$ orbit of the hydrogen atom is:

Similar Questions

The ratio of the areas of the electron orbits for the second excited state to the first excited state for the hydrogen atom is

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. What is the ratio of the kinetic energy to the potential energy of the electron in this state?

The de Broglie wavelength of an electron revolving in the first Bohr orbit of circumference $2\pi r$ is ............

The angular momentum of an electron in Bohr's hydrogen atom having energy $(-0.544) \text{ eV}$ is ($h$ = Planck's constant)

$A$ hydrogen atom in the ground state absorbs $ 10.2 \text{ eV} $ of energy. The orbital angular momentum of the electron is increased by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module