Find the value of wave number (bar{v}) in ter | vedclass

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(B) The Rydberg formula is given by $\bar{v} = \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_2 > n_1$.Given tha

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$\frac{32R}{9}$

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(B) The Rydberg formula is given by $\bar{v} = \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $n_2 > n_1$.Given that the sum of the energy levels is $n_2 + n_1 = 4$ and their difference is $n_2 - n_1 = 2$.Adding these two equations: $2n_2 = 6 \implies n_2 = 3$.Subtracting the equations: $2n_1 = 2 \implies n_1 = 1$.For the $He^{+}$ ion,the atomic number $Z = 2$.Substituting these values into the Rydberg formula:$\bar{v} = R 	imes (2)^2 \left( \frac{1}{1^2} - \frac{1}{3^2} ight)$$\bar{v} = R 	imes 4 \left( 1 - \frac{1}{9} ight)$$\bar{v} = 4R \left( \frac{8}{9} ight) = \frac{32R}{9}$.Thus,the correct option is $B$.

Find the value of wave number $(\bar{v})$ in terms of Rydberg's constant,when the transition of an electron takes place between two levels of $He^{+}$ ion whose sum is $4$ and difference is $2$.

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