To excite the spectral line of wavelength 496 | vedclass

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(D) The energy of the excited state is given by the sum of the ground state energy and the excitation energy.$E_{	ext{excited}} = E_{	ext{ground}} +

Vedclass · Accepted Answer

$15.7 	ext{ eV}, 18.2 	ext{ eV}$

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(D) The energy of the excited state is given by the sum of the ground state energy and the excitation energy.$E_{	ext{excited}} = E_{	ext{ground}} + E_{	ext{excitation}} = 10.5 	ext{ eV} + 7.7 	ext{ eV} = 18.2 	ext{ eV}$.This represents the higher energy level $(E_1)$ involved in the transition.The energy of the emitted photon corresponding to the wavelength $\lambda = 4960 \mathring A = 496 	ext{ nm}$ is calculated as:$E_{	ext{photon}} = \frac{hc}{\lambda} = \frac{1240 	ext{ eV nm}}{496 	ext{ nm}} = 2.5 	ext{ eV}$.Since the photon is emitted during a transition from a higher level $(E_1)$ to a lower level $(E_2)$, we have:$E_{	ext{photon}} = E_1 - E_2$$2.5 	ext{ eV} = 18.2 	ext{ eV} - E_2$$E_2 = 18.2 	ext{ eV} - 2.5 	ext{ eV} = 15.7 	ext{ eV}$.Thus, the energies of the two levels are $18.2 	ext{ eV}$ and $15.7 	ext{ eV}$.

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