Some of the wavelengths observed in the emiss | vedclass

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(D) The energy of a photon with wavelength $\lambda$ is given by $E = \frac{12400 \, 	ext{eV} \cdot \mathring{A}}{\lambda (\mathring{A})}$.Calculatin

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$3646$

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(D) The energy of a photon with wavelength $\lambda$ is given by $E = \frac{12400 \, 	ext{eV} \cdot \mathring{A}}{\lambda (\mathring{A})}$.Calculating the energies of the given wavelengths:$E_1 = \frac{12400}{912} \approx 13.6 \, 	ext{eV}$$E_2 = \frac{12400}{1026} \approx 12.08 \, 	ext{eV}$$E_3 = \frac{12400}{1216} \approx 10.2 \, 	ext{eV}$$E_4 = \frac{12400}{3646} \approx 3.4 \, 	ext{eV}$At room temperature,hydrogen atoms are in the ground state $(n=1)$.For absorption to occur,the photon energy must match the energy difference between the ground state and an excited state $(n > 1)$. The minimum energy required for excitation is $E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, 	ext{eV}$.Since $3.4 \, 	ext{eV}$ is less than the required $10.2 \, 	ext{eV}$ for excitation from the ground state,the wavelength $3646 \, \mathring{A}$ will not be strongly absorbed.

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