Atoms & Nuclei

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16 of 239

Question 16 of 239

In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5A∘.0.5A^{^{\circ}}. If the speed of electron is 2.2×106ms−1.2.2\times {10}^{6}m{s}^{-1}. Then the current associated with the electron will be__________ ×10−2mA.\times {10}^{-2}mA. [Take π\pi as 227]\left.\dfrac{22}{7}\right]

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Answer: 112

Time taken to revolve in circular orbit is T=2πω=2πrv=2×227×5×10−112.2×106=107×10−16sT=\dfrac{2\pi }{\omega }=\dfrac{2\pi r}{v}=\dfrac{2\times \dfrac{22}{7}\times 5\times {10}^{-11}}{2.2\times {10}^{6}}=\dfrac{10}{7}\times {10}^{-16}s
Now, current associated with the electron is I=eT=1.6×10−19107×10−16=1.12×10−3AI=\dfrac{e}{T}=\dfrac{1.6\times {10}^{-19}}{\dfrac{10}{7}\times {10}^{-16}}=1.12\times {10}^{-3}A
⇒I=1.12mA\Rightarrow I=1.12mA
Hence, the current is 112×102mA112\times {10}^{2}mA.

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