91Ó°ÊÓ

The rate at which energy is emitted from an acceleration charge q and the acceleration a is $\frac{\mathbf{dE}}{\mathbf{dt}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{6}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{o}}{\mathbf{c}}^{\mathbf{3}}}$q is the acceleration charge, a is the acceleration of the hydrogen atom, ${\mathit{Î\mu }}_{\mathbf{0}}$is the electric constant, c is the speed of the light. The acceleration of the circular orbit is given as $\mathbf{a}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$v is the velocity of the circular orbit, R is the radius of the circular orbit

The acceleration of the circular orbit is $\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$Multiply numerator and denominator with $\frac{1}{2}m$we get,$\mathrm{a}=\frac{{\displaystyle \frac{1}{2}}{\mathrm{mv}}^2}{{\displaystyle \frac{1}{2}}\mathrm{mR}}$Kinetic energy of the electron is ${\mathrm{K}}_{\mathrm{E}}=13.6\mathrm{eV}$. Substitute the values

$$\begin{gathered} {\mathrm{K}}_{\mathrm{E}}=13.6\mathrm{eV}×1.6×{10}^{-19}\mathrm{J}/\mathrm{ev} \\ =2.176×{10}^{-18}\mathrm{J} \end{gathered}$$

Substitute ${\mathrm{K}}_{\mathrm{E}}$as $\frac{1}{2}\mathrm{m}$in$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$ we get,

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{K}}_{\mathrm{E}}}{{\displaystyle \frac{1}{2}}\mathrm{mR}} \\ =\frac{2{\mathrm{K}}_{\mathrm{E}}}{\mathrm{mR}} \end{gathered}$$

Substitute the values we have,

$$\begin{gathered} \mathrm{a}=\frac{2\left(2.176×{10}^{-18}\mathrm{J}\right)}{\left(9.109×{10}^{-31}\mathrm{kg}\right)\left(0.059×{10}^{-9}\mathrm{m}\right)} \\ =9.031×{10}^{22}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the circular orbit is$9.031×{10}^{22}\mathrm{m}/{\mathrm{s}}^2$

The rate at which energy is emitted from an acceleration charge q and the acceleration a is$\frac{\mathrm{dE}}{\mathrm{dt}}=\frac{{\mathrm{q}}^2{\mathrm{a}}^2}{6{\mathrm{Ï„Ï„Î\mu }}_{\mathrm{o}}{\mathrm{c}}^3}$Substitute the values we get,

$$\begin{gathered} \frac{\mathrm{dE}}{\mathrm{dt}}=\frac{{\left(1×{10}^{-19}\right)}^{2}9.031×{10}^{22}}{6\mathrm{π}\left(8.854×{10}^{-12}\right)\left(2.998×{10}^8\right)} \\ =4.64×{10}^{-8}\mathrm{Nm}/\mathrm{s} \\ =4.64×{10}^{-8}\mathrm{J}/\mathrm{s} \end{gathered}$$

To convert energy in electron volt into energy in joules is${\mathrm{E}}_{\mathrm{ev}}={\mathrm{E}}_{\mathrm{s}}=\left(6.241×{10}^{18}\right)$Substitute the values we get,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{ev}}=4.64×{10}^{-8}\mathrm{J}/\mathrm{s}\left(6.241×{10}^{18}\right) \\ =2.89×{10}^{11}\mathrm{ev}/\mathrm{s} \end{gathered}$$

Thus, the energy required for the hydrogen atom is$2.89×{10}^{11}\mathrm{ev}/\mathrm{s}$

The fraction of its energy radiates per second is

$\frac{{\displaystyle \frac{\mathrm{dE}}{\mathrm{dt}}}\left(1\mathrm{s}\right)}{\mathrm{E}}=\frac{{\mathrm{E}}_{\mathrm{hydrogen}}}{{\mathrm{K}}_{\mathrm{v}}}$

Substitute the values we get,

$$\begin{gathered} \frac{{\displaystyle \frac{\mathrm{dE}}{\mathrm{dt}}}\left(1\mathrm{s}\right)}{\mathrm{E}}=\frac{2.89×{10}^{11}}{13.6\mathrm{ev}} \\ =2.125×{10}^{10}\mathrm{per}\sec \end{gathered}$$

The value of rate of energy emission $\frac{dE}{dt}$is larger value, it means that the electrons in hydrogen atom almost losses it all energy instantly.

Thus, the energy required for the hydrogen atom in radiate per second is $2.125×{10}^{10}\mathrm{per}\sec$