JEE MAIN - Physics (2011 - No. 2)
Energy required for the electron excitation in $$L{i^{ + + }}$$ from the first to the third Bohr orbit is :
$$36.3$$ $$eV$$
$$108.8$$ $$eV$$
$$122.4$$ $$eV$$
$$12.1$$ $$eV$$
Explanation
Energy of excitation,
$$\Delta E = 13.6\,{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)eV$$
$$ \Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {{1 \over {{1^2}}} - {1 \over {{3^2}}}} \right) = 108.8\,eV$$
$$\Delta E = 13.6\,{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)eV$$
$$ \Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {{1 \over {{1^2}}} - {1 \over {{3^2}}}} \right) = 108.8\,eV$$
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