question_answer

Find the number of photons of light with wavelength 4000pm that provide U of energy.

Answer:

Number of photon\[(n)=?\] \[\lambda =4000\text{ }pm=4000\times {{10}^{12}}m\] Energy \[(E)=1J\] \[E=nhv\] \[\Rightarrow \]\[E=\frac{nhc}{\lambda }\]                                         \[\left( \because v=\frac{C}{\lambda } \right)\] \[\Rightarrow \]\[n=\frac{E\lambda }{hC}\]where h is Planck?s constant with a value of\[6.625\times {{10}^{34}}Js\]and C is velocity of light in air or vacuum (\[3\times {{10}^{8}}m/s\]) \[n=\frac{1\times 4000\times {{10}^{-12}}}{6.625\times {{10}^{-34}}\times 3\times {{10}^{8}}}=2\times {{10}^{16}}\] Therefore, an approximate of\[2\times {{10}^{16}}\]photons provides the energy of 1J.