When the electron in the hydrogen atom jumps | vedclass

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.F

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$\frac{27}{32}\lambda$

Vedclass · Answer

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first transition ($n_2 = 2$ to $n_1 = 1$):$\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4}$.Thus,$R = \frac{4}{3\lambda}$.For the second transition ($n_2 = 3$ to $n_1 = 1$):$\frac{1}{\lambda'} = R \left[ \frac{1}{1^2} - \frac{1}{3^2} ight] = R \left[ 1 - \frac{1}{9} ight] = \frac{8R}{9}$.Substituting the value of $R$:$\frac{1}{\lambda'} = \left( \frac{4}{3\lambda} ight) 	imes \frac{8}{9} = \frac{32}{27\lambda}$.Therefore,$\lambda' = \frac{27}{32}\lambda$.

When the electron in the hydrogen atom jumps from the $2^{nd}$ orbit to the $1^{st}$ orbit,the wavelength of the radiation emitted is $\lambda$. When the electron jumps from the $3^{rd}$ orbit to the $1^{st}$ orbit,the wavelength of the emitted radiation will be:

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