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\[\overset{o}{\mathop{A}}\,\]), 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: |
| \[hc=\] (Plancks constant) velocity of light) |
| \[=\frac{(6.6\times {{10}^{-34}}J-s)\,(3\times {{10}^{8}}\,m/s)}{(1.6\times {{10}^{-19}}\,J/eV)}\] |
| \[=12.375\times {{10}^{-7}}\,\,eV-m=12375\,eV-\overset{o}{\mathop{A}}\,\] |
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