First slide

Bohr's atom model

Question

Consider a hydrogen like atom whose energy in n^th excited state is given by $E n = - \frac{13.6 Z^{2}}{n^{2}}$ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy $E_{max} = 52.2224 eV$ and least energetic photons have energy $E_{min} = 1.224 eV$ . What is the atomic number of atom?

Moderate

Solution

Maximum energy is liberated for transition $E_{n} \to E_{1}$ and minimum energy for $E_{n} \to E_{n - 1}$
$Hence \frac{E_{1}}{n^{2}} - E_{1} = 52.224 eV and \frac{E_{1}}{n^{2}} - \frac{E_{1}}{(n - 1)^{2}} = 1.224 eV$
$\begin{array}{rcl} \Rightarrow & E_{1} = - 54.4 eV and n = 5 \\ Now E_{1} & = & - \frac{13.6 Z^{2}}{1^{2}} = - 54.4 eV . Hence Z = 2 \end{array}$

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