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Tests for Hydrogen

If the value ...

Question

If the value of $E_{n}$ $=$ 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
$E_{n} =$ 78.4 kcal/mole => $78.4 \times 4.2 =$ 329.28 kJ/mole $= \frac{- 329.28}{96.5}$ eV $=$ 3.4 eV, which is the energy of 2nd orbit of Hydrogen atom.

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