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Structure Of Atom

Question
CBSEENCH11005276

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate: (a) threshold wavelength and (b) Planck’s constant.
λ(nm)                           500             450              400
v x 10-5(cm s-1)             2.55            4.35               5.35

Solution

Let the threshold wavelength  = λ0nm
                                          = λx 10-9m
Then straight h left parenthesis straight v minus straight v subscript 0 right parenthesis space equals space 1 half mv squared
  or   hc open parentheses 1 over straight lambda minus 1 over straight lambda subscript 0 close parentheses space equals space 1 half mv squared
For the first experiment:
                   hc over 10 to the power of negative 9 end exponent open parentheses 1 over 500 minus 1 over straight lambda subscript 0 close parentheses space equals space 1 half straight m left parenthesis 2.55 space cross times 10 to the power of 6 right parenthesis squared
 For seconds experiment:
hc over 10 to the power of negative 9 end exponent open parentheses 1 over 450 minus 1 over straight lambda subscript 0 close parentheses equals space 1 half straight m left parenthesis 4.35 cross times 10 to the power of 6 right parenthesis squared space space space space... left parenthesis 2 right parenthesis
For the third experiment:
 hc over 10 to the power of negative 9 end exponent open parentheses 1 over 400 minus 1 over straight lambda subscript 0 close parentheses space equals space 1 half straight m space left parenthesis 5.20 space cross times space 10 to the power of 6 right parenthesis squared
                                                                 ...(3)

Dividing equation (2) by equation (1), we have
      fraction numerator straight lambda subscript 0 minus 450 over denominator 450 space straight lambda subscript 0 end fraction cross times space fraction numerator 500 space straight lambda subscript 0 over denominator straight lambda subscript 0 minus 500 end fraction space equals space open parentheses fraction numerator 4.35 over denominator 2.55 end fraction close parentheses squared
or space space space fraction numerator straight lambda subscript 0 minus 450 over denominator straight lambda subscript 0 minus 500 end fraction space equals space 450 over 500 open parentheses fraction numerator 4.35 over denominator 2.55 end fraction close parentheses squared space equals space 2.619
or space space straight lambda subscript 0 minus 450 space equals space 2.619 space straight lambda subscript 0 space minus space 1309.5
or space space space 1.619 space straight Å space equals space 859.5 space space space space space space space therefore space space space straight lambda subscript 0 space equals space 531 space nm

 Substituting this value in equation (3), we have
fraction numerator straight h cross times left parenthesis 3 cross times 10 to the power of 8 right parenthesis over denominator 10 to the power of negative 9 end exponent end fraction space open parentheses 1 over 400 minus 1 over 531 close parentheses
space equals space 1 half left parenthesis 9.11 space cross times space 10 to the power of negative 31 end exponent right parenthesis space left parenthesis 5.20 space cross times 10 to the power of 6 right parenthesis squared
space equals space 6.66 space cross times space 10 to the power of negative 34 end exponent Js