Question 27

An electron of a hydrogen like atom, having $$Z = 4$$, jumps from $$4^{th}$$ energy state to $$2^{nd}$$ energy state. The energy released in this process, will be: (Given $$Rch = 13.6$$ eV)
Where $$R =$$ Rydberg constant, $$c =$$ Speed of light in vacuum, $$h =$$ Planck's constant

Using Bohr's energy formula: $$\Delta E = Rch \cdot Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$

$$\Delta E = 13.6 \times (4)^2 \times \left(\frac{1}{2^2} - \frac{1}{4^2}\right)$$

$$\Delta E = 13.6 \times 16 \times \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times 16 \times \frac{3}{16}$$

$$\Delta E = 13.6 \times 3 = 40.8\text{ eV}$$

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