When an electron makes a transition from (n+1 | vedclass

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(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = n$ and $n_2 = n+1$.Subst

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$v = \frac{2cR Z^2}{n^3}$

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(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = n$ and $n_2 = n+1$.Substituting the values: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$.Simplifying the expression: $\frac{1}{\lambda} = R Z^2 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = R Z^2 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.Given the condition $n >> 1$,we can approximate $2n+1 \approx 2n$ and $(n+1)^2 \approx n^2$.Thus,$\frac{1}{\lambda} \approx R Z^2 \left( \frac{2n}{n^2 \cdot n^2} \right) = \frac{2R Z^2}{n^3}$.Since $v = \frac{c}{\lambda}$,we have $v = \frac{2cR Z^2}{n^3}$.

When an electron makes a transition from $(n+1)$ state to $n^{th}$ state,the frequency of emitted radiation is related to $n$ according to $(n >> 1)$:

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