Physics-
General
Easy

Question

The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of3

  1. 1 0 to the power of 20 end exponent    
  2. 1 0 to the power of 19 end exponent    
  3. 1 0 to the power of 17 end exponent    
  4. 1 0 to the power of 15 end exponent    

The correct answer is: 1 0 to the power of 15 end exponent


    m v r equals fraction numerator h over denominator 2 pi end fraction (for first orbit)
    rightwards double arrow m omega r to the power of 2 end exponent equals fraction numerator h over denominator 2 pi end fraction rightwards double arrow m cross times 2 pi nu cross times r to the power of 2 end exponent equals fraction numerator h over denominator 2 pi end fraction rightwards double arrow nu equals fraction numerator h over denominator 4 pi to the power of 2 end exponent m r to the power of 2 end exponent end fraction
    equals fraction numerator 6.6 cross times 1 0 to the power of negative 34 end exponent over denominator 4 left parenthesis 3.14 right parenthesis to the power of 2 end exponent cross times 9.1 cross times 1 0 to the power of negative 31 end exponent cross times left parenthesis 0.53 cross times 1 0 to the power of negative 10 end exponent right parenthesis to the power of 2 end exponent end fraction equals 6.5 cross times 1 0 to the power of 15 end exponent fraction numerator r e v over denominator s e c end fraction

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