Assuming the atom is in the ground state,the | vedclass

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

Question

(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{

Vedclass · Accepted Answer

$\frac{\mu_0 e^7 \pi m^2}{8 \varepsilon_0^3 h^5}$

Vedclass · Answer

(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{e^2}{4\pi\varepsilon_0 r}$.From Bohr's quantization condition: $mvr = \frac{nh}{2\pi}$. For ground state $(n=1)$,$mvr = \frac{h}{2\pi} \implies v = \frac{h}{2\pi mr}$.Substituting $v$ into the force equation: $m(\frac{h}{2\pi mr})^2 = \frac{e^2}{4\pi\varepsilon_0 r} \implies \frac{h^2}{4\pi^2 mr^2} = \frac{e^2}{4\pi\varepsilon_0 r} \implies r = \frac{\varepsilon_0 h^2}{\pi m e^2}$.The current $I$ due to the orbiting electron is $I = \frac{e}{T} = \frac{ev}{2\pi r}$.The magnetic field at the center is $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.Substituting $v = \frac{h}{2\pi mr}$ and $r = \frac{\varepsilon_0 h^2}{\pi m e^2}$:$B = \frac{\mu_0 e}{4\pi r^2} \cdot \frac{h}{2\pi mr} = \frac{\mu_0 e h}{8\pi^2 m r^3}$.Substituting $r^3 = (\frac{\varepsilon_0 h^2}{\pi m e^2})^3 = \frac{\varepsilon_0^3 h^6}{\pi^3 m^3 e^6}$:$B = \frac{\mu_0 e h}{8\pi^2 m} \cdot \frac{\pi^3 m^3 e^6}{\varepsilon_0^3 h^6} = \frac{\mu_0 e^7 \pi m^2}{8 \varepsilon_0^3 h^5}$.

Similar Questions

An atom of hydrogen gas is in an excited state $n_2$. It absorbs a photon of some energy and jumps to a higher energy state $n_1$. Then,it returns to the ground state after emitting six different wavelengths in the emission spectrum. Determine the values of $n_1$ and $n_2$.

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

What is the minimum energy that must be given to a $H$ atom in the ground state so that it can emit an $H_{\gamma}$ line in the Balmer series? If the angular momentum of the system is conserved,what would be the angular momentum of such an $H_{\gamma}$ photon?

The velocity of an electron in the second orbit of a sodium atom (atomic number $Z = 11$) is $v$. The velocity of an electron in its fifth orbit will be

$A$ stationary hydrogen atom undergoes a transition from $n=5$ to $n=4$. The recoil speed of the atom is (where $R=$ Rydberg constant,$h=$ Planck's constant,and $m=$ mass of the hydrogen atom).

Vedclass Test Series

Exam Paper Generator

Online Exam Module