Question #827e8

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Question #827e8

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Answer 1 · 2017-04-20T02:28:09

$4.09 \cdot 10^{- 19} J$ $4.09 * 10^(-19)J$

Explanation:

Use the formula:
$E_{n} = - R_{H} (\frac{1}{n^{2}})$ $E_n = -R_H(1/n^2)$
where :
$R_{H} = Rydberg constant = 2.18 \cdot 10^{- 18} J$ $R_H = "Rydberg constant" = 2.18 * 10^(-18)J$ ,
$n = the energy level$ $n = "the energy level"$

For $n = 4$ $n = 4$ :
$E_{4} = - 2.18 \cdot 10^{- 18} J (\frac{1}{4^{2}}) = - 1.3625 \cdot 10^{- 19} J$ $E_4 = -2.18 * 10^(-18)J(1/4^2) = -1.3625 *10^(-19)J$

For $n = 2$ $n = 2$
$E_{2} = - 2.18 \cdot 10^{- 18} J (\frac{1}{2^{2}}) = - 5.45 \cdot 10^{- 19} J$ $E_2 = -2.18 * 10^(-18)J(1/2^2) = -5.45 *10^(-19)J$

Then to find the energy of the photon, you just calculate the change in energy.

$DeltaE = E_"initial" - E_"final"$

$DeltaE = (-1.3625 *10^(-19)J) - (-5.45 *10^(-19)J)$

$DeltaE = 4.0875 * 10^(-19)J$

Since you only have 3 significant figures, the answer would be:
$4.09 * 10^(-19)J$

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Question #827e8

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