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From the given data calculate the number of revolutions of an electron in the second Bohr orbit in one second |
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Answer» Solution :If `r_(2)` is the radius of the second Bohr orbit, the distance travelled by an electron in one revolution will be `2pi r_(2)` (i.e., the circumference). We have calculate that an electron travels a distance of `1.09 xx 10^(8)cm` in one second in the second Bohr orbit. Hence, revolution per second `=(v_(2))/(2pi r_(2))= (1.09 xx 10^(8))/(2pi r_(2))` Now `r_(2)= 2^(2). r_(1)` `=2^(2) xx 53 xx 10^(-8)` `=2.12 xx 10^(-8)` (n=2, `r_(1)= 0.53 xx 10^(-8)`) `therefore` revolutions per second `= (1.09 xx 10^(8))/(2 xx (3.14) xx 2.12 xx 10^(-8))` `=8.18 xx 10^(14)` |
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