Based on the equation E = -2.178 times 10^{-1 | vedclass

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(C) The energy of an electron at an infinite distance from the nucleus is defined as zero. As an electron approaches the nucleus,the electrostatic att

Vedclass · Accepted Answer

For $n = 1$,the electron has a more negative energy than it does for $n = 6$,which means that the electron is more loosely bound in the smallest allowed orbit.

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(C) The energy of an electron at an infinite distance from the nucleus is defined as zero. As an electron approaches the nucleus,the electrostatic attraction increases,causing the energy of the electron to decrease and become negative. Therefore,as the value of $n$ decreases (i.e.,the orbit is closer to the nucleus),the energy of the electron becomes more negative,meaning it is more tightly bound,not loosely bound. Thus,the statement in option $C$ is incorrect because it claims the electron is more loosely bound in the smallest orbit.

Based on the equation $E = -2.178 \times 10^{-18} \ J \left( \frac{Z^2}{n^2} \right)$,certain conclusions are written. Which of them is not correct?

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