The maximum wavelength that a sample of hydro | vedclass

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(B) Hydrogen atoms in their ground state $(n=1)$ can absorb photons to transition to higher energy levels $(n=2, 3, 4, \dots)$.To obtain the maximum w

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$1216 \,\mathring{A}$

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(B) Hydrogen atoms in their ground state $(n=1)$ can absorb photons to transition to higher energy levels $(n=2, 3, 4, \dots)$.To obtain the maximum wavelength $(\lambda_{\max})$,we must consider the transition with the minimum energy difference,which is from $n=1$ to $n=2$.The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Substituting $n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.Thus,$\lambda_{\max} = \frac{4}{3R}$.Given $R \approx 1.097 	imes 10^7 \, 	ext{m}^{-1}$,we have $\lambda_{\max} = \frac{4}{3 	imes 1.097 	imes 10^7} \approx 1.216 	imes 10^{-7} \, 	ext{m} = 1216 \, \mathring{A}$.

The maximum wavelength that a sample of hydrogen atoms can absorb is

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