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(D) The energy of the photon is given by $E = \frac{hc}{\lambda}$.Substituting the given values:$E = \frac{4 	imes 10^{-15} \ eV \cdot s 	imes 3 	i

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Electron will keep orbiting in the ground state of the atom

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(D) The energy of the photon is given by $E = \frac{hc}{\lambda}$.Substituting the given values:$E = \frac{4 	imes 10^{-15} \ eV \cdot s 	imes 3 	imes 10^{8} \ m/s}{300 	imes 10^{-9} \ m} = \frac{12 	imes 10^{-7}}{300 	imes 10^{-9}} \ eV = \frac{1200}{300} \ eV = 4 \ eV$.The energy required to excite a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is $\Delta E = 13.6 \ eV 	imes (1 - \frac{1}{4}) = 13.6 	imes 0.75 = 10.2 \ eV$.Since the energy of the photon $(4 \ eV)$ is less than the energy required for the first excitation $(10.2 \ eV)$ and also less than the ionization energy $(13.6 \ eV)$, the electron cannot absorb this energy to transition to a higher state.Therefore, the electron will remain in the ground state.

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