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  • Highly energetic electrons are bombaarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is <img alt="14^{\frac{1}{3}}" src="https://

Highly energetic electrons are bombaarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is 14^{\frac{1}{3}}. The atomic number of nucleus will be

Option: 1

25


Option: 2

26


Option: 3

56


Option: 4

30


Answers (1)

best_answer

Since \mathrm{R\, \alpha\, A^{\frac{1}{3}}}.

\mathrm{\Rightarrow \frac{R}{R_{H e}}=\left(\frac{A}{4}\right)^{1 / 3} \Rightarrow\left(\frac{A}{4}\right)^{1 / 3}=(14)^{1 / 3}}
\mathrm{\Rightarrow \frac{A}{4}=14 \Rightarrow A=56 }.

Since \mathrm{A=Z+N \Rightarrow 56=Z+30 }
\mathrm{\Rightarrow Z=26 }.

Posted by

Ritika Harsh

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