A) \[4125\,\,nm\]
B) \[412.5\,\,nm\]
C) \[41.250\,\,nm\]
D) \[4\,\,nm\]
Correct Answer: B
Solution :
Energy of photon,\[E=\frac{hc}{\lambda }\] If energy\[E\]is expressed in\[eV\]and wavelength \[\lambda \](in\[\overset{\text{o}}{\mathop{\text{A}}}\,\]), then \[E=\frac{12375}{\lambda (\overset{\text{o}}{\mathop{\text{A}}}\,)}eV\] \[\therefore \] \[\lambda =\frac{12375}{E}\overset{\text{o}}{\mathop{\text{A}}}\,=\frac{12375}{3}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=4125\overset{\text{o}}{\mathop{\text{A}}}\,=412.5\,\,nm\]
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