An electron, $\mathrm{e}_{1}$ is moving in the fifth stationary state, and another electron $\mathrm{e}_{2}$…

An electron, $\mathrm{e}_{1}$ is moving in the fifth stationary state, and another electron $\mathrm{e}_{2}$ is moving in the fourth stationary state. The radius of orbit of electron, $\mathrm{e}_{1}$ is five times the radius of orbit of electron, $\mathrm{e}_{2}$ calculate the ratio of velocity of electron $\mathrm{e}_{1}\left(\mathrm{v}_{1}ight)$ to the velocity of electron $\mathrm{e}_{2}\left(\mathrm{v}_{2}ight)$
  1. $5: 1$
  2. $4: 1$
  3. $1: 5$
  4. $1: 4$

Solution

From the expression of Bohr's theory, we know that
$m_{e} \mathrm{v}_{1} r_{1}=n_{1} \frac{h}{2 \pi}$
$\& m_{e} \mathrm{v}_{2} r_{2}=n_{2} \frac{h}{2 \pi}$
$\frac{m_{e} \mathrm{v}_{1} r_{1}}{m_{e} \mathrm{v}_{2} r_{2}}=\frac{n_{1}}{n_{2}} \frac{h}{2 \pi} \times \frac{2 \pi}{h}$
Given, $r_{1}=5 r_{2}, n_{1}=5, n_{2}=4$
$\frac{m_{e} \times \mathrm{v}_{1} \times 5 r_{2}}{m_{e} \times \mathrm{v}_{2} \times r_{2}}=\frac{5}{4}$
$\Rightarrow \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{5}{4 \times 5}=\frac{1}{4}=1: 4$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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