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s344n2d4d5 [400]
3 years ago
7

HURRYYYY pls help due today n i be giving brainliest!!! n no mfkn links plsss

Physics
1 answer:
nordsb [41]3 years ago
5 0

Answer:

1 Frequency

2 Wavelength

3 Amplitude

4 Crest

Hope it helps pls mark brainliest

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Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
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The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

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Energy of the second orbit in H atom .

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Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

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During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

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E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

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