A) \[-5.44\times {{10}^{-19}}eV\]
B) \[-5.44\times {{10}^{-19}}cal\]
C) \[-5.44\times {{10}^{-19}}kJ\]
D) \[-5.44\times {{10}^{-19}}J\]
Correct Answer: D
Solution :
\[{{E}_{n}}=\frac{13.6\times {{Z}^{2}}}{{{n}^{2}}}eV\] Given, \[n=2\] For hydrogen, \[Z=1\] \[\therefore \] \[{{E}_{2}}=-\frac{13.6\times {{(1)}^{2}}}{{{(2)}^{2}}}eV\] \[=-\frac{13.6}{4}eV\] \[=-\frac{(13.6\times 1.6\times {{10}^{-19}})}{4}J\] \[=-5.44\times {{10}^{-19}}\,J\]
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