A) \[E=12.4\,h\nu \]
B) \[E=12.4\,h/\lambda \]
C) \[E=12.4\,/\lambda \]
D) \[E=h\nu \]
Correct Answer: C
Solution :
Energy of photon \[E=\frac{hc}{\lambda }\] (Joules) \[=\frac{hc}{e\lambda }(eV)\] \[\Rightarrow \,\underset{(eV)}{\mathop{E}}\,=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times \lambda ({\AA})}=\frac{12375}{\lambda ({\AA})}\] \[\Rightarrow \,\,E(keV)=\frac{12.37}{\lambda ({\AA})}\approx \frac{12.4}{\lambda }\]
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