A) \[-5.44\times {{10}^{-19}}J\]
B) \[-5.44\times {{10}^{-19}}\,kJ\]
C) \[-5.44\times {{10}^{-19}}\,cal\]
D) \[-5.44\times {{10}^{-19}}\,eV\]
Correct Answer: A
Solution :
\[E=-\frac{2.172\times {{10}^{-18}}}{{{n}^{2}}}=\frac{-2.172\times {{10}^{-18}}}{{{2}^{2}}}\] \[=-5.42\times {{10}^{-19}}J\].You need to login to perform this action.
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