Question 494: Atoms & Nuclei

Question

Using Bohr's model, the orbital period of electron in hydrogen atom in the $$\mathrm{n}^{	ext {th }}$$ orbit is $$\left(\varepsilon_0=ight.$$ permittivity of vacuum, $$\mathrm{h}=$$ Planck's constant, $$\mathrm{m}=$$ mass of electron, $$\mathrm{e}=$$ electronic charge)

Accepted Answer

The period of an electron in the $$\mathrm{n}^{	ext {th }}$$ orbit of a hydrogen atom is given by
$$\begin{aligned}
& \mathrm{T}_{\mathrm{n}}=\frac{2 \pi \mathrm{r}_{\mathrm{n}}}{\mathrm{v}_{\mathrm{n}}} \
& 	ext { and } \mathrm{r}_{\mathrm{n}}=\frac{\varepsilon_0 \mathrm{n}^2 \mathrm{~h}^2}{\pi \mathrm{me}^2} \
& 	ext { and linear speed } \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{nh}} \
& 	herefore \mathrm{T}_{\mathrm{n}}=\frac{2 \pi \cdot \varepsilon_0 \mathrm{n}^2 \mathrm{~h}^2}{\pi \mathrm{me}^2} 	imes \frac{2 \varepsilon_0 \mathrm{nh}}{\mathrm{e}^2}=\frac{4 \varepsilon_0^2 \mathrm{n}^3 \mathrm{~h}^3}{\mathrm{me}^4}
\end{aligned}$$

Answer

$$\frac{4 \varepsilon_0 \mathrm{nh}^3}{\mathrm{me}^2}$$

Answer

$$\frac{4 \varepsilon_0 \mathrm{n}^2 \mathrm{~h}^2}{\mathrm{me}^2}$$

Answer

$$\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}$$

Answer

$$\frac{4 \varepsilon_0^2 \mathrm{n}^2 \mathrm{~h}^3}{m \mathrm{e}^3}$$

Atoms & Nuclei

Using Bohr's model, the orbital period of electron in hydrogen atom in the $\mathrm{n}^{\text {th }} orbit is \left(\varepsilon_0=\right. permittivity of vacuum, \mathrm{h}= Planck's constant, \mathrm{m}= mass of electron, \mathrm{e}=$ electronic charge)

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