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Question

If the value of En = 78.4 kcal/mole, then the order of the orbit in hydrogen atom is:

A
2
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B
3
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C
4
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D
1
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Solution

The correct option is A 2
En= 78.4 kcal/mole => 78.4×4.2= 329.28 kJ/mole =329.2896.5 eV = 3.4 eV, which is the energy of 2nd orbit of Hydrogen atom.

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