An electron jumps from the 4^{text{th}} orbit | vedclass

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(B) Concept: The wavelength $\lambda$ of electromagnetic radiation emitted is given by the Rydberg formula:$\frac{1}{\lambda} = R_{H} \left( \frac{1}{

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

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(B) Concept: The wavelength $\lambda$ of electromagnetic radiation emitted is given by the Rydberg formula:$\frac{1}{\lambda} = R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \quad \dots(1)$The frequency $f$ of the emitted radiation is related to wavelength by:$f = \frac{c}{\lambda} \quad \dots(2)$Substituting equation $(1)$ into equation $(2)$,we get:$f = c R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$Given values: $c = 3 	imes 10^8 \ m/s$,$R_{H} = 10^7 \ m^{-1}$,$n_1 = 2$,and $n_2 = 4$.Substituting these values:$f = (3 	imes 10^8) 	imes 10^7 	imes \left( \frac{1}{2^2} - \frac{1}{4^2} ight)$$f = 3 	imes 10^{15} 	imes \left( \frac{1}{4} - \frac{1}{16} ight)$$f = 3 	imes 10^{15} 	imes \left( \frac{4-1}{16} ight)$$f = 3 	imes 10^{15} 	imes \frac{3}{16}$$f = \frac{9}{16} 	imes 10^{15} \ Hz$

An electron jumps from the $4^{\text{th}}$ orbit to the $2^{\text{nd}}$ orbit of a hydrogen atom. Given Rydberg's constant $R_{H}=10^7 \ m^{-1}$,calculate the frequency in $Hz$ of the emitted radiation. (Take $c=3 \times 10^8 \ m/s$)

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