Physics-
General
Easy

Question

In a photoelectric experiment the relation between applied potential difference between cathode and anode V and the photoelectric current I was found to be shown in graph below. If Planck’s constant h equals 6.6 cross times 10 to the power of negative 34 end exponent J s, the frequency of incident radiation would be nearly left parenthesis i n blank s to the power of negative 1 end exponent right parenthesis

  1. 0.436 cross times 10 to the power of 18 end exponent    
  2. 0.436 cross times 10 to the power of 17 end exponent    
  3. 0.775 cross times 10 to the power of 15 end exponent    
  4. 0.775 cross times 10 to the power of 16 end exponent    

The correct answer is: 0.775 cross times 10 to the power of 15 end exponent


    For photoelectric effect,
    e V subscript 0 end subscript equals h v rightwards double arrow v equals fraction numerator e V subscript 0 end subscript over denominator h end fraction
    v equals fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 3.2 over denominator 6.6 cross times 10 to the power of negative 34 end exponent end fraction
    equals 0.775 cross times 10 to the power of 15 end exponent H z

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