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  4. If the electron falls from n=5 to n=4 in the H-atom, then emitted energy is

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Q. If the electron falls from $n=5$ to $n=4$ in the H-atom, then emitted energy is

Structure of Atom

Solution:

$E_{n}=-\frac{13.6}{n^{2}} eV$

$E_{5}-E_{4}=13.6\left(\frac{1}{4^{2}}-\frac{1}{5^{2}}\right) eV$

$ =13.6\left(\frac{1}{16}-\frac{1}{25}\right)=0.306\, eV$