A) \[Z=6\]
B) \[Z=4\]
C) \[Z=11\]
D) \[Z=44\]
Correct Answer: A
Solution :
\[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\,=\frac{{{({{Z}_{2}}-1)}^{2}}}{{{({{Z}_{1}}-1)}^{2}}}\,\,\,\,\left[ as\,\frac{1}{\lambda }\propto \,{{(Z-1)}^{2}} \right]\] \[\frac{1}{4}=\,\frac{{{({{Z}_{2}}-1)}^{2}}}{{{(11-1)}^{2}}}\]
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