CBSE Class 11-science Answered
9?
Asked by smanishkumar2002 | 20 May, 2018, 22:45: PM
Question:
Suppose 10-17 J energy is needed by interior human eye to see an object . How many photons of green light (wavelenght =550 nm) are needed to generate this minimum amount of energy?
Solution:
![Given colon
straight E equals space 10 to the power of negative 17 space end exponent straight J
straight lambda space equals 550 space nm
straight lambda equals space 550 cross times 10 to the power of negative 9 end exponent space straight m.
We space have comma
straight E space equals space hc over straight lambda
space space space space space equals fraction numerator 6.62510 to the power of negative 34 end exponent space cross times space 3 cross times 10 to the power of 8 over denominator 550 cross times 10 to the power of negative 9 end exponent end fraction
straight E space equals space 3.61 space cross times 10 to the power of negative 19 end exponent space straight J
No space. of space photons space required space equals fraction numerator 10 to the power of negative 17 space end exponent over denominator 3.61 space cross times 10 to the power of negative 19 end exponent end fraction
No space. of space photons space required space equals space 2.77 cross times 10 squared
No space. of space photons space required space almost equal to space space 28 space photons.](https://images.topperlearning.com/topper/tinymce/cache/1aa840ba402ce4402a1813a40f2914bf.png)
Answered by Varsha | 21 May, 2018, 10:33: AM
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