A) \[3.08\times {{10}^{15}}{{s}^{-1}}\]
B) \[2.00\times {{10}^{15}}{{s}^{-1}}\]
C) \[1.54\times {{10}^{15}}{{s}^{-1}}\]
D) \[1.03\times {{10}^{15}}{{s}^{-1}}\]
Correct Answer: A
Solution :
\[{{E}_{\text{ionisation}}}={{E}_{\infty }}-{{E}_{n}}=\frac{13.6Z_{eff}^{2}}{{{n}^{2}}}eV\] = \[\left[ \frac{13.6{{Z}^{2}}}{n_{2}^{2}}-\frac{13.6{{Z}^{2}}}{n_{1}^{2}} \right]\] \[E=h\nu =\frac{13.6\times {{1}^{2}}}{{{(1)}^{2}}}-\frac{13.6\times {{1}^{2}}}{{{(4)}^{2}}}\]; \[h\nu =13.6-0.85\] \[\because \] \[h=6.625\times {{10}^{-34}}\] \[\nu =\frac{13.6-0.85}{6.625\times {{10}^{-34}}}\times 1.6\times {{10}^{-19}}\] = \[3.08\times {{10}^{15}}{{s}^{-1}}\].You need to login to perform this action.
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