A) 4.2V
B) 3.5V
C) 2.1V
D) none of these
Correct Answer: C
Solution :
The energy of first excitation of sodium is \[E=hv=\frac{hc}{\lambda }\] where h is Plancks constants, v is frequency, c is speed of light and \[\lambda \] is wavelength. \[\therefore \] \[E=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5896\times {{10}^{10}}}J\] \[E=3.37\times {{10}^{-19}}J\] Also, since \[1.6\times {{10}^{-19}}J=1\,eV\] \[\therefore \] \[E=\frac{3.37\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[E=2.1\,eV\] Hence, corresponding first excitation potential is 2.1V.
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