A) \[2.00\]
B) \[2.82\]
C) \[4.00\]
D) \[1.414\]
Correct Answer: D
Solution :
Given, the maximum electron density in the morning \[{{N}_{\max }}\,={{10}^{10}}\,{{m}^{-3}}\] \[f_{e}^{'}\,=9\,{{({{N}_{\max }})}^{1/2}}\] \[=9{{(10)}^{1/2}}\] Critical frequency, \[f_{e}^{'}\,=9\times {{10}^{5}}\,Hz\] The maximum electron density at noon \[{{N}_{\max }}\,=2\times {{10}^{10}}\,{{m}^{-3}}\] Critical frequency, \[f_{e}^{''}\,=9({{N}_{\max }}){{\,}^{1/2}}\] \[=9\,{{(\,2\times {{10}^{10}})}^{1/2}}\] \[=9\times \sqrt{2}\,\times {{10}^{2}}\] The ratio of critical frequency at noon and critical frequency in the morning \[\frac{f_{e}^{''}}{{{f}_{e}}}\,=\frac{9\times \sqrt{2}\times {{10}^{5}}}{9\times {{10}^{5}}}\] \[\frac{f_{e}^{''}}{f_{e}^{'}}=\sqrt{2}=1.414\]
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