A) \[{{10}^{19}}\]
B) \[2\times {{10}^{20}}\]
C) 60
D) zero
Correct Answer: B
Solution :
Key Idea: Energy of all the photons emitted per second is the power of the bulb. Energy per second or power of bulb is given by \[P=n\,h\,v\] but \[v=\frac{c}{\lambda }\] \[\therefore \] \[P=\frac{nhc}{\lambda }\] or \[n=\frac{P\lambda }{hc}\] ?(i) Here \[P=50\,W,\lambda =600\,nm=600\times {{10}^{-9}}m\] \[h=6.6\times {{10}^{-34}}\,J-s,\,c=3\,\times {{10}^{8}}\,m/s\] Substituting the values in Eq. (i), we find \[n=\frac{50\times 600\times {{10}^{-9}}}{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}\] \[\approx \,\,2\times {{10}^{20}}\]
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