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Atoms and Nuclei question

2021 · 27 Jul · Shift 1 · Q68
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  5. /2021 · 27 Jul · Shift 1 · Q68

Atoms and Nuclei question

2021 · 27 Jul · Shift 1 · Q68

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 Ao\mathop A\limits^oAo​. If the speed of electron is 2.2 ×\times× 166 m/s, then the current associated with the electron will be ‾\underline{\hspace{2cm}}​×\times× 10 −-− 2 mA. [Take π\piπ as 227{{22} \over 7}722​]
Numerical answer
View written solutionFree

Correct answer: 112

  1. Formula for current due to revolving electron

If an electron revolves in a circular orbit, the current is

I=eTI = \frac{e}{T}I=Te​

where TTT is the time period of revolution.

Since

T=2πrvT = \frac{2\pi r}{v}T=v2πr​

therefore,

I=ev2πrI = \frac{e v}{2\pi r}I=2πrev​

  1. Given data
  • Radius of orbit: r=0.5 A˚=0.5×10−10 m=5×10−11 mr = 0.5\,\mathring{A} = 0.5 \times 10^{-10}\,\text{m} = 5 \times 10^{-11}\,\text{m}r=0.5A˚=0.5×10−10m=5×10−11m
  • Speed of electron: The printed value clearly intends the standard Bohr speed, v=2.2×106 m/sv = 2.2 \times 10^6\,\text{m/s}v=2.2×106m/s
  • Charge of electron: e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C
  • Take π=227\pi = \frac{22}{7}π=722​
  1. Substitute into the formula

I=(1.6×10−19)(2.2×106)2×227×5×10−11I = \frac{(1.6 \times 10^{-19})(2.2 \times 10^6)}{2 \times \frac{22}{7} \times 5 \times 10^{-11}}I=2×722​×5×10−11(1.6×10−19)(2.2×106)​

First, numerator:

1.6×2.2=3.521.6 \times 2.2 = 3.521.6×2.2=3.52

so

numerator=3.52×10−13\text{numerator} = 3.52 \times 10^{-13}numerator=3.52×10−13

Now denominator:

2×227×5×10−11=447×5×10−112 \times \frac{22}{7} \times 5 \times 10^{-11} = \frac{44}{7} \times 5 \times 10^{-11}2×722​×5×10−11=744​×5×10−11

=2207×10−11= \frac{220}{7} \times 10^{-11}=7220​×10−11

Hence,

I=3.52×10−132207×10−11I = \frac{3.52 \times 10^{-13}}{\frac{220}{7} \times 10^{-11}}I=7220​×10−113.52×10−13​

I=3.52×10−13×7220×1011I = 3.52 \times 10^{-13} \times \frac{7}{220} \times 10^{11}I=3.52×10−13×2207​×1011

I=3.52×7220×10−2I = \frac{3.52 \times 7}{220} \times 10^{-2}I=2203.52×7​×10−2

I=24.64220×10−2I = \frac{24.64}{220} \times 10^{-2}I=22024.64​×10−2

I=0.112×10−2 AI = 0.112 \times 10^{-2}\,\text{A}I=0.112×10−2A

  1. Convert to mA

Since

1 A=103 mA1\,\text{A} = 10^3\,\text{mA}1A=103mA

we get

I=0.112×10−2×103 mAI = 0.112 \times 10^{-2} \times 10^3\,\text{mA}I=0.112×10−2×103mA

I=1.12 mAI = 1.12\,\text{mA}I=1.12mA

Now write it in the asked form:

1.12 mA=112×10−2 mA1.12\,\text{mA} = 112 \times 10^{-2}\,\text{mA}1.12mA=112×10−2mA

  1. Final answer

The required number is

112\boxed{112}112​

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