Consider a hydrogen like atom whose energy in nth excited state is given by $$En=$$−$$13.6Z2n2$$ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy $$Emax=52.2224eV$$ and least energetic photons have energy $$Emin=1.224eV$$. What is the atomic number of atom?

Question

Consider a hydrogen like atom whose energy in nth excited state is given by $$En=$$−$$13.6Z2n2$$ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy $$Emax=52.2224eV$$ and least energetic photons have energy $$Emin=1.224eV$$. What is the atomic number of atom?

Infinity Learn · Accepted Answer

Maximum energy is liberated for transition $E_{n} \to E_{1}$ and minimum energy for $E_{n} \to E_{n - 1}$

$Hence \frac{E_{1}}{n^{2}} - E_{1} = 52.224 eV and \frac{E_{1}}{n^{2}} - \frac{E_{1}}{(n - 1)^{2}} = 1.224 eV$

$\begin{array}{rcl} \Rightarrow & E_{1} = - 54.4 eV and n = 5 \\ Now E_{1} & = & - \frac{13.6 Z^{2}}{1^{2}} = - 54.4 eV . Hence Z = 2 \end{array}$

Bohr's atom model

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Bohr's atom model

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Maximum energy is liberated for transition En→E1 and minimum energy for En→En−1 Hence E1n2−E1=52.224eV and E1n2−E1(n−1)2=1.224eV ⇒ E1=−54.4eV and n=5 Now E1=−13.6Z212=−54.4eV. Hence Z=2

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