SATHEE: Structure of Atom - Result Question 14

Structure of Atom - Result Question 14

####14. The energy of second Bohr orbit of the hydrogen atom is $- 328 k J m o l^{- 1}$ ; hence the energy of fourth Bohr orbit would be:

[2005]

(a) $- 41 k J m o l^{- 1}$

(b) $- 82 k J m o l^{- 1}$

(c) $- 164 k J m o l^{- 1}$

(d) $- 1312 k J m o l^{- 1}$

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Solution:

(b) $E_{n} = E_{1} \times (\frac{Z}{n})^{2}$

$E_{2} = \frac{E_{1}}{2^{2}} = - 328 k J / m o l$

$E_{4} = \frac{E_{1}}{4^{2}}$

$\Rightarrow \frac{E_{4}}{E_{2}} = \frac{2^{2}}{4^{2}} = \frac{1}{2^{2}}$

$\Rightarrow E_{4} = \frac{E_{2}}{2_{2}} = \frac{- 328}{4} k J / m o l = - 82 k j / m o l$

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