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

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(NONE) The equation $E = -2.178 	imes 10^{-18} \ J \ \frac{Z^2}{n^2}$ describes the energy of an electron in a hydrogen-like atom.Option $A$ is corre

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 tightly bound in the smallest allowed orbit.

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(NONE) The equation $E = -2.178 	imes 10^{-18} \ J \ \frac{Z^2}{n^2}$ describes the energy of an electron in a hydrogen-like atom.Option $A$ is correct because as $n$ increases,energy becomes less negative (closer to zero),meaning the electron is less tightly bound.Option $B$ is correct as the negative sign indicates a stable bound state relative to the zero-energy state at infinity.Option $C$ is correct because the radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z}$,so $r \propto n^2$.Option $D$ is correct as $\Delta E = E_f - E_i$ can be calculated using this formula.Since all statements $A, B, C,$ and $D$ are scientifically correct based on the Bohr model,there is no incorrect statement provided in the options.

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

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