Answer:
Number of photons \[(n)=3\times {{10}^{18}}\] Energy \[(E)=1.5\text{ }J\] Wavelength\[(\lambda )~=?\] \[h=6.63\times {{10}^{34}}JS\] \[E=\frac{nhc}{\lambda }\] \[\Rightarrow \] \[\lambda =\frac{nhc}{E}=\frac{3\times {{10}^{18}}\times 6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.5}\] \[=39.78\times {{10}^{8}}m\](or) \[3978\times {{10}^{10}}m=3978\overset{{}^\circ }{\mathop{\text{A}}}\,\]
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