Calculate the two longest wavelengths of the radiation emitted when hydrogen atoms make transitions from higher states to $n = 2$ state.

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Solution

According to balmer formula;
Wave no = $R_{H} [1 / {n_{1}}^{2} - 1 / {n_{2}}^{2}]$
Here $n_{1} = 2$ , $n_{2} = 4$ , $R_{H}$ =109678
Putting these values in the equation we get,
Wave number =109678 $(1 / 2^{2} - 1 / 4^{2})$ =109678 X 3/16
Also $λ$ =1/wave number
Therefore λ = 16 / (109678 $\times$ 3) = 486 nm.

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Calculate the two longest wavelengths of the radiation emitted when hydrogen atoms make transitions from higher states to n=2 state.

According to balmer formula;Wave no = RH[1/ n12− 1/ n22]Here n1=2, n2=4, RH=109678Putting these values in the equation we get,Wave number =109678 (1/22−1/42) =109678 X 3/16Also λ=1/wave numberTherefore λ = 16 / (109678 × 3) = 486 nm.

Calculate the two longest wavelengths of the radiation emitted when hydrogen atoms make transitions from higher states to $n = 2$ state.