A) \[\frac{25}{3}\times {{10}^{19}}J,0.4\,A\]
B) \[\frac{25}{4}\times {{10}^{19}}J,6.2\,A\]
C) \[\frac{25}{2}\times {{10}^{19}}J,0.8\,A\]
D) None of the above
Correct Answer: A
Solution :
From Einsteins relation, the energy of photon of wavelength \[(\lambda )\] is \[E=\frac{hc}{\lambda }\] Given, \[h=6.6\times {{10}^{-34}}Js\], \[c=3\times {{10}^{8}}m/s\], \[\lambda =6600\,\overset{o}{\mathop{A}}\,\] \[\lambda =6600\times {{10}^{-10}}m\] \[\therefore \] \[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6600\times {{10}^{-10}}}\] \[=3\times {{10}^{-19}}J\] Number of photons emitted per second by 25 W source is \[n=\frac{25\,W\times 1\,\,s}{3\times {{10}^{-19}}}\] \[=\frac{25}{3}\times {{10}^{19}}\] and \[I=ne\] \[=\frac{3}{100}\times \frac{25}{3}\times {{10}^{19}}\times 1.6\times {{10}^{-19}}\] = 0.4 A
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