A) 1.4 eV
B) 4.9 eV
C) 3.1 eV
D) 1.6 eV
Correct Answer: C
Solution :
From Plancks quantum theory, light travels in the form of small bundles or packets of energy called photons. The energy of each photon is \[E=hv=\frac{hc}{\lambda }\] where h is Plancks constant, v is frequency, c is speed of light and \[\lambda \], is wavelength. Given, \[{{\lambda }_{1}}=6000\,\overset{o}{\mathop{A}}\,,\,{{E}_{1}}=3.32\times {{10}^{-19}}J\], \[{{\lambda }_{2}}=4000\,\overset{o}{\mathop{A}}\,\] \[\therefore \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}\] \[\Rightarrow \] \[\frac{3.32\times {{10}^{19}}}{{{E}_{2}}}=\frac{4000}{6000}\] \[\Rightarrow \] \[{{E}_{2}}=\frac{6}{4}\times 3.32\times {{10}^{-19}}J\] \[=4.98\times {{10}^{-19}}J\] \[=\frac{4.98\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}\,eV=3.1\,eV\]
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