In hydrogen and hydrogen-like atoms,the ratio | vedclass

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(D) The energy of an electron in the $n$-th orbit of a hydrogen-like atom is given by $E_n = -\frac{Z^2 E_0}{n^2}$,where $E_0 = 13.6 	ext{ eV}$.Calcu

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$Z^0 n^0$

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(D) The energy of an electron in the $n$-th orbit of a hydrogen-like atom is given by $E_n = -\frac{Z^2 E_0}{n^2}$,where $E_0 = 13.6 	ext{ eV}$.Calculating the energy differences:$E_{4n} - E_{2n} = -\frac{Z^2 E_0}{(4n)^2} - \left( -\frac{Z^2 E_0}{(2n)^2} ight) = Z^2 E_0 \left( \frac{1}{4n^2} - \frac{1}{16n^2} ight) = Z^2 E_0 \left( \frac{4-1}{16n^2} ight) = \frac{3 Z^2 E_0}{16n^2}$.$E_{2n} - E_n = -\frac{Z^2 E_0}{(2n)^2} - \left( -\frac{Z^2 E_0}{n^2} ight) = Z^2 E_0 \left( \frac{1}{n^2} - \frac{1}{4n^2} ight) = Z^2 E_0 \left( \frac{4-1}{4n^2} ight) = \frac{3 Z^2 E_0}{4n^2}$.Taking the ratio:$\frac{E_{4n} - E_{2n}}{E_{2n} - E_n} = \frac{3 Z^2 E_0 / 16n^2}{3 Z^2 E_0 / 4n^2} = \frac{4}{16} = \frac{1}{4}$.Since the result is a constant,it is independent of $Z$ and $n$,which can be expressed as $Z^0 n^0$.

In hydrogen and hydrogen-like atoms,the ratio of the difference of energies $E_{4n}-E_{2n}$ and $E_{2n}-E_{n}$ varies with its atomic number $Z$ and $n$ as:

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