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The energy of the electron in the second and third Bohr’s orbit of the hydrogen atom is $- 5.42 \times 10^{- 12}$ erg and $- 2.42 \times 10^{- 12}$ erg respectively calculate the wavelength of the emitted light (in cm). When the electron drops from the third to the second orbit with multiplying of $10^{5}$ .

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answer is 6.62.

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Detailed Solution

plan $Δ E = {E_{n_{2}}}_{} - {E_{n_{1}}}_{}$

Also $Δ E = h υ = \frac{h c}{λ}$

Thurs $λ = \frac{h c}{Δ E}$

$Δ E = (E_{3} - E_{2}) = - 2.42 \times 10^{- 12} - (- 5.42 \times 10^{- 12})$

$= 3.0 \times 10^{- 12}$ ergs

$λ = \frac{6.62 \times 10^{- 27} e r g s \times 3 \times 10^{10} c m s^{- 1}}{3.0 \times 10^{- 12} e r g s} = 6.62 \times 10^{- 5} c m$

$= 6.62 \times 10^{- 5} \times 10^{5}$

$= 6.62$

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