Taking Rydberg's constant R_H = 1.097 times 1 | vedclass

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(D) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the Ba

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$6563 \ \mathring{A}, 4861 \ \mathring{A}$

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(D) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Balmer series,$n_1 = 2$.For the first wavelength (first line),$n_2 = 3$:$\frac{1}{\lambda_1} = 1.097 	imes 10^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = 1.097 	imes 10^7 \left[ \frac{1}{4} - \frac{1}{9} ight] = 1.097 	imes 10^7 \left[ \frac{5}{36} ight]$.$\lambda_1 = \frac{36}{5 	imes 1.097 	imes 10^7} \approx 6.563 	imes 10^{-7} \ m = 6563 \ \mathring{A}$.For the second wavelength (second line),$n_2 = 4$:$\frac{1}{\lambda_2} = 1.097 	imes 10^7 \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = 1.097 	imes 10^7 \left[ \frac{1}{4} - \frac{1}{16} ight] = 1.097 	imes 10^7 \left[ \frac{3}{16} ight]$.$\lambda_2 = \frac{16}{3 	imes 1.097 	imes 10^7} \approx 4.861 	imes 10^{-7} \ m = 4861 \ \mathring{A}$.

Taking Rydberg's constant $R_H = 1.097 \times 10^7 \ m^{-1}$,the first and second wavelengths of the Balmer series in the hydrogen spectrum are:

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