If m is the mass of an electron,v is its velo | vedclass

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

Question

(A) In the $C.G.S.$ system,the electrostatic force of attraction between the nucleus (charge $Ze$) and the electron (charge $e$) provides the necessar

Vedclass · Accepted Answer

$\frac{1}{2} \frac{Ze^2}{r}$

Vedclass · Answer

(A) In the $C.G.S.$ system,the electrostatic force of attraction between the nucleus (charge $Ze$) and the electron (charge $e$) provides the necessary centripetal force for circular motion.The electrostatic force is given by $F_e = \frac{(Ze)(e)}{r^2} = \frac{Ze^2}{r^2}$.The centripetal force required for circular motion is $F_c = \frac{mv^2}{r}$.Equating the two forces: $\frac{mv^2}{r} = \frac{Ze^2}{r^2}$.Multiplying both sides by $r$,we get $mv^2 = \frac{Ze^2}{r}$.The kinetic energy $K$ is defined as $K = \frac{1}{2}mv^2$.Substituting the expression for $mv^2$: $K = \frac{1}{2} \left( \frac{Ze^2}{r} \right) = \frac{Ze^2}{2r}$.

If $m$ is the mass of an electron,$v$ is its velocity,and $r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's postulate,the kinetic energy $K = \frac{1}{2}mv^2$ of the electron in the $C.G.S.$ system is equal to:

Similar Questions

The ionization potential of a hydrogen-like atom is $122.4 \, V$. Find its atomic number $Z$.

The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is

When a hydrogen atom absorbs a photon of wavelength $60 \ nm$,the atom undergoes photo-ionization. What will be the maximum kinetic energy of the emitted electron in $eV$?

If an electron jumps from the $1^{st}$ orbital to the $3^{rd}$ orbital,then it will:

If in a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is ${r_n}$ and the frequency of revolution of the electron in the ${n^{th}}$ orbit is ${f_n}$,choose the correct option.

Vedclass Test Series

Exam Paper Generator

Online Exam Module