In Bohr’s model, the electron revolves around the nucleus in a circular path of radius 5.1×10−11m at a frequency of 6.8×1015revolutions per second. The equivalent current is

i=qt=net =3.8×1015×1.6×10−19=10.88×10−4=1.088×10−3A=1.09×10−3A

In Bohr’s model, the electron revolves around the nucleus in a circular path of radius 5.1×10−11m at a frequency of 6.8×1015revolutions per second. The equivalent current is

i=qt=net =3.8×1015×1.6×10−19=10.88×10−4=1.088×10−3A=1.09×10−3A

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In Bohr’s model, the electron revolves around the nucleus in a circular path of radius $5 .1 \times 10^{- 11} m$ at a frequency of $6 {.8\times10}^{15}$ revolutions per second. The equivalent current is

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$1 .09 \times 10^{- 3} A$

$2 \times 10^{- 3} A$

$1 .5 \times 10^{- 3} A$

$0 .5 \times 10^{- 3} A$

answer is A.

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$i = \frac{q}{t} = \frac{ne}{t} \begin{array}{l} = 3 .8 \times 10^{15} \times 1 .6 \times 10^{- 19} \\ = 10 .88 \times 10^{- 4} \\ = 1 .088 \times 10^{- 3} A \\ = 1 .09 \times 10^{- 3} A \end{array}$

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In Bohr’s model, the electron revolves around the nucleus in a circular path of radius 5.1×10−11m at a frequency of 6.8×1015revolutions per second. The equivalent current is

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In Bohr’s model, the electron revolves around the nucleus in a circular path of radius $5 .1 \times 10^{- 11} m$ at a frequency of $6 {.8\times10}^{15}$ revolutions per second. The equivalent current is