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Structure Of Atom

Question
CBSEENCH11005266

In astronomical observations, signals observed from the distant starts are generally weak. If the photon detector receives a total of 3.15 x 10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

Solution

Energy (E) of one photon = hv = straight h straight c over straight lambda            ...(1)
Here straight h space equals space 6.626 space cross times space 10 to the power of negative 34 end exponent Js semicolon space space space straight c space equals space 3 space cross times space 10 to the power of 8 space ms to the power of negative 1 end exponent semicolon
space
    straight lambda space equals space 600 space nm space equals space 600 space cross times space 10 to the power of negative 9 end exponent straight m
Substituting the values  in eq. (1), we have
  straight E space equals fraction numerator left parenthesis 6.626 space cross times 10 to the power of negative 34 end exponent space Js right parenthesis space cross times space left parenthesis 3 cross times 10 to the power of 8 space ms to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 600 space cross times space 10 to the power of negative 9 end exponent straight m right parenthesis end fraction
space space equals space 3.313 space cross times space 10 to the power of negative 19 end exponent straight J
Total space energy space received space equals space 3.15 space cross times space 10 to the power of negative 18 end exponent straight J
therefore space space Number space of space photons space received
space space space space space space space space space space space space space space equals fraction numerator 3.15 space cross times space 10 to the power of negative 18 end exponent over denominator 3.313 space cross times space 10 to the power of negative 19 end exponent straight J end fraction space equals 9.51 space equals space 10