Based on equation E = - 2.178 times 10^{-18} | vedclass

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(B) The energy of an electron in a hydrogen-like atom is given by $E = - 2.178 	imes 10^{-18} \ J \left( \frac{Z^2}{n^2} ight)$.As $n$ increases,th

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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(B) The energy of an electron in a hydrogen-like atom is given by $E = - 2.178 	imes 10^{-18} \ J \left( \frac{Z^2}{n^2} ight)$.As $n$ increases,the energy becomes less negative (closer to zero).For $n = 1$,the energy is $-2.178 	imes 10^{-18} \ J$,and for $n = 6$,it is $-2.178 	imes 10^{-18} \ J 	imes (1/36)$,which is a smaller negative value.More negative energy implies that the electron is more strongly bound to the nucleus.Therefore,the statement in option $B$ is incorrect because it claims the electron is more loosely bound in the smallest orbit $(n=1)$,whereas it is actually more strongly bound.

Based on 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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