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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.

Similar Questions

Consider a hydrogen atom with its electron in the $n^{\text{th}}$ orbital. An electromagnetic radiation of wavelength $90 \ nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 \ eV$,then the value of $n$ is $(hc = 1242 \ eV \ nm)$.

The ratio of the total energy of the $2^{\text{nd}}$ orbit electron for the hydrogen atom $(_1H^1)$ to that of the helium ion $(He^+)$ $(_2^4He)$ is:

For an electron moving in the $n^{\text{th}}$ Bohr orbit,the de Broglie wavelength of the electron is:

When an electron in a hydrogen atom jumps from the third excited state to the ground state,the de-Broglie wavelength associated with the electron becomes

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