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The potential energy of a particle of mass m is given by U left parenthesis x right parenthesis equals open curly brackets table row cell E subscript 0 end subscript semicolon       0 less or equal than x less or equal than 1 end cell row cell 0   semicolon       x greater than 1 end cell end table close lambda1 and lambda 2 are the de-Broglie wavelengths of the particle, when 0 less or equal than x less or equal than 1 and x > 1 respectively. If the total energy of particle is 2E0, the ratio fraction numerator lambda subscript 1 end subscript over denominator lambda subscript 2 end subscript end fraction will be

  1. 2    
  2. 1    
  3. square root of 2    
  4. fraction numerator 1 over denominator square root of 2 end fraction    

The correct answer is: square root of 2


    K.E.= 2 E0– E0 = E0 (for 0 £ x £ 1) Þ lambda subscript 1 end subscript equals fraction numerator h over denominator square root of 2 m E subscript 0 end subscript end root end fraction
    K.E. = 2 E0 (for x > 1) Þ lambda subscript 2 end subscript equals fraction numerator h over denominator square root of 4 m E subscript 0 end subscript end root end fraction rightwards double arrow fraction numerator lambda subscript 1 end subscript over denominator lambda subscript 2 end subscript end fraction equals square root of 2.

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