If the value of $E_{n} = - 78.4 k c a l m o l^{- 1}$ , the order of the orbit in hydrogen atom is:

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Solution

The correct option is C 2
$E_{n} = - 78.4 k c a l m o l^{- 1}$
$= - 78.4 \times 4.2 k J m o l^{- 1}$
$= - 329.28 k J m o l^{- 1}$
$= - \frac{329.28}{96.5} e V = - 3.4 e V$
$- 3, 4 e V = - 13.6 \times \frac{Z^{2}}{n^{2}} \Rightarrow - 3, 4 e V = - 13.6 \times \frac{1}{n^{2}} \Rightarrow n^{2} = 4 \Rightarrow n = 2$
This energy corresponds to $2^{n d}$ orbit of H atom.

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Similar questions

E_{n} = - 78.4

E_{n} = - 78.4 k c a l m o l^{- 1}

E_{n}

=

n^{t h}

E_{n} = - \frac{313.6}{n^{2}}

E_{n} = - 313.6 / n^{2}

k c a l / m o l

E = - 34.84

k c a l / m o l

n

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