A) 4125 nm
B) 412.5 nm
C) 41250 nm
D) 4 nm
Correct Answer: B
Solution :
If energy E is expressed in eV and wavelength \[\lambda \] (in \[\overset{\text{o}}{\mathop{\text{A}}}\,\]), then energy of photon, \[E=\frac{12375\,}{\lambda \,({\AA})}eV\] \[\therefore \lambda =\frac{12375}{E\,(eV)}{\AA}\] \[=\frac{12375\,}{3\,eV}{\AA}\,-4125\,{\AA}=412.5\,nm\] Note: Energy of photon is \[E=\frac{hc}{\lambda (\overset{o}{\mathop{\text{A}}}\,)}=\frac{12375}{\lambda (\overset{o}{\mathop{\text{A}}}\,)}eV\] Here, hc = 12375 comes from the following procedure: hc = (Planck?s constant) velocity of light) \[=\frac{(6.6\times {{10}^{-34}}J-s)\,(3\times {{10}^{8}}\,m/s)}{(1.6\times {{10}^{-19}}\,J/eV)}\] \[=12.375\times {{10}^{-7}}\,\,eV-m=12375\,eV-\overset{\text{0}}{\mathop{\text{A}}}\,\]
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