Q.The energy of an electron in the $n^{th}$ Bohr orbit of Hydrogen atom is given by $E_n = \frac{-13.6}{n^2} \text{ eV}$ . The energy of a photon emitted when an electron jumps from $n=3$ to $n=2$ is:

1.89 \text{ eV}

3.40 \text{ eV}

1.51 \text{ eV}

12.09 \text{ eV}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.The energy of an electron in the nthn^{th}nth Bohr orbit of Hydrogen atom is given by En=−13.6n2 eVE_n = \frac{-13.6}{n^2} \text{ eV}En​=n2−13.6​ eV. The energy of a photon emitted when an electron jumps from n=3n=3n=3 to n=2n=2n=2 is: