Question 451: Atoms & Nuclei

Question

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

Accepted Answer

$\mathrm{r}=$ radius of $\mathrm{n}^{\mathrm{th}}$ orbit $=\frac{\mathrm{n}^2 \mathrm{~h}^2}{\pi \mathrm{mZe}^2}$
$\mathrm{v}=$ speed of $\mathrm{e}^{-}$in $\mathrm{n}^{	ext {th }}$ orbit $=\frac{\mathrm{Ze}^2}{2 \varepsilon_0 n h}$
$\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$
Substituting values of $v$ and $r$,
$	herefore \quad \mathrm{T}=\frac{4 \varepsilon_0^2 \mathrm{n}^3 \mathrm{~h}^3}{\mathrm{mZ} \mathrm{Z}^2 \mathrm{e}^4}$
For hydrogen, $Z=1$
$	herefore \quad \mathrm{T}=\frac{4 \varepsilon_0^2 \mathrm{n}^3 \mathrm{~h}^3}{\mathrm{me}^4}$

Answer

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

Answer

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

Answer

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

Answer

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

Atoms & Nuclei

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

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