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(B) In a standard hydrogen atom,the potential energy is $U = -\frac{27.2 \ eV}{n^2}$ and the total energy is $E = -\frac{13.6 \ eV}{n^2}$.If we shift

Vedclass · Accepted Answer

The total energy of the shell decreases with an increase in the value of $n$.

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(B) In a standard hydrogen atom,the potential energy is $U = -\frac{27.2 \ eV}{n^2}$ and the total energy is $E = -\frac{13.6 \ eV}{n^2}$.If we shift the reference level such that the potential energy at the ground state $(n=1)$ is zero,we add a constant $C$ to the potential energy expression.Since $U(1) = -27.2 + C = 0$,we get $C = 27.2 \ eV$.Thus,the new potential energy is $U' = -\frac{27.2}{n^2} + 27.2 \ eV$.The total energy $E'$ becomes $E' = K + U' = \frac{13.6}{n^2} + (-\frac{27.2}{n^2} + 27.2) = 27.2 - \frac{13.6}{n^2} \ eV$.For $n=1$,$E' = 27.2 - 13.6 = 13.6 \ eV$.As $n$ increases,$\frac{13.6}{n^2}$ decreases,so $E'$ increases. Therefore,statement $B$ is incorrect.

In the hydrogen atom,if the reference level of potential energy is assumed to be zero at the ground state level,choose the incorrect statement.

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