An electron jumps from the 4th orbit to the 2 | vedclass

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(C) The Rydberg formula for the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Given $n_1

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$\frac{9}{16} 	imes 10^{15}$

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(C) The Rydberg formula for the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Given $n_1 = 2$,$n_2 = 4$,and $R = 10^5 	ext{ cm}^{-1} = 10^7 	ext{ m}^{-1}$.$\frac{1}{\lambda} = 10^5 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 10^5 \left( \frac{1}{4} - \frac{1}{16} ight) = 10^5 \left( \frac{3}{16} ight) 	ext{ cm}^{-1}$.Thus,$\lambda = \frac{16}{3} 	imes 10^{-5} 	ext{ cm} = \frac{16}{3} 	imes 10^{-7} 	ext{ m}$.The frequency $
u$ is given by $
u = \frac{c}{\lambda}$,where $c = 3 	imes 10^8 	ext{ m/s}$.$
u = \frac{3 	imes 10^8}{\frac{16}{3} 	imes 10^{-7}} = \frac{9}{16} 	imes 10^{15} 	ext{ Hz}$.

An electron jumps from the $4th$ orbit to the $2nd$ orbit of a hydrogen atom. Given the Rydberg constant $R = 10^5 \text{ cm}^{-1}$. The frequency in $\text{Hz}$ of the emitted radiation will be:

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