Question 670: Atoms & Nuclei

Question

Consider a hydrogen-like ionized atom with atomic number $$Z$$ with a single electron. In the emission spectrum of this atom, the photon emitted in the $$n=2$$ to $$n=1$$ transition has energy $$74.8eV$$ higher than the photon emitted in the $$n=3$$ to $$n=2$$ transition. The ionization energy of the hydrogen atom is $$13.6$$ $$eV.$$ The value of $$Z$$ is ____________.

Accepted Answer

We know that,
$$\Delta {E_{i - j}} = 13.6\left[ {{1 \over {n_j^2}} - {1 \over {n_i^2}}} ight]{Z^2}$$
Therefore,
$$\Delta {E_{2 - 1}} = 13.6\left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} ight]{Z^2} = 13.6\left[ {{1 \over 1} - {1 \over 4}} ight]{Z^2}$$
$$\Delta {E_{2 - 1}} = 13.6 	imes {3 \over 4} 	imes {Z^2}$$
Similarly,
$$\Delta {E_{3 - 2}} = 13.6\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight]{Z^2} = 13.6\left[ {{1 \over 4} - {1 \over 9}} ight]{Z^2}$$
$$\Delta {E_{3 - 2}} = 13.6 	imes {5 \over {36}} 	imes {Z^2}$$
Given, $$\Delta {E_{2 - 1}} = \Delta {E_{3 - 2}} + 74.8$$
$$ \Rightarrow 13.6 	imes {3 \over 4} 	imes {Z^2} = 13.6 	imes {5 \over {36}} 	imes {Z^2} + 74.8$$
$$ \Rightarrow 13.6 	imes {3 \over 4} 	imes {Z^2} - 13.6 	imes {5 \over {36}} 	imes {Z^2} = 74.8$$
$$ \Rightarrow 13.6 	imes {Z^2}\left[ {{3 \over 4} - {5 \over {36}}} ight] = 74.8$$
$$ \Rightarrow 13.6 	imes {Z^2}\left[ {{{27 - 5} \over {36}}} ight] = 74.8$$
$$ \Rightarrow 13.6 	imes {Z^2} 	imes {{22} \over {36}} = 74.8$$
$$ \Rightarrow {Z^2} = {{74.8 	imes 36} \over {13.6 	imes 22}} = 9$$
$$ \Rightarrow Z = 3$$

Atoms & Nuclei

Integer / Numeric Answer Type