In $$hydrogen-like$$ atom (atomic$$ number A is in a higher excited state of quantum number n. This excited atom can make a transition to first excited state by emitting photons of energies $$10.20$$ eV and $$17.00$$ eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $$4.25$$ eV and $$5.95$$ eV respectively. Determine values of n $$(ionization$$ energy of hydrogen atom $$=$$ $$13.6$$ $$eV).

Question

Infinity Learn · Accepted Answer

Given that En−$$E2=10.2+17.0=27.2eV$$ and En−$$E3=4.25+5.95=10.2eV$$⇒$$E3$$−$$E2=27.2$$−$$10.2=17eV$$⇒$$1.89Z2=17eV$$⇒$$Z=3$$ Now we use $$27.2=13.6Z214$$−$$1n2$$⇒$$n=6$$

Bohr's atom model

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$Given that E_{n} - E_{2} = 10.2 + 17.0 = 27.2 eV$
$and E_{n} - E_{3} = 4.25 + 5.95 = 10.2 eV$
$\Rightarrow E_{3} - E_{2} = 27.2 - 10.2 = 17 eV$
$\Rightarrow 1.89 Z^{2} = 17 eV$
$\Rightarrow Z = 3$
$Now we use 27.2 = 13.6 Z^{2} (\frac{1}{4} - \frac{1}{n^{2}}) \Rightarrow n = 6$

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Bohr's atom model

Remember concepts with our Masterclasses.

Given that En−E2=10.2+17.0=27.2eV and En−E3=4.25+5.95=10.2eV⇒E3−E2=27.2−10.2=17eV⇒1.89Z2=17eV⇒Z=3 Now we use 27.2=13.6Z214−1n2⇒n=6

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$Given that E_{n} - E_{2} = 10.2 + 17.0 = 27.2 eV$
$and E_{n} - E_{3} = 4.25 + 5.95 = 10.2 eV$
$\Rightarrow E_{3} - E_{2} = 27.2 - 10.2 = 17 eV$
$\Rightarrow 1.89 Z^{2} = 17 eV$
$\Rightarrow Z = 3$
$Now we use 27.2 = 13.6 Z^{2} (\frac{1}{4} - \frac{1}{n^{2}}) \Rightarrow n = 6$