The second ionization energy (the energy required to removethe second outermost

Question

The second ionization energy (the energy required to removethe second outermost electron) of calcium is 11.9 eV. Determine themaximum wavelength of incident radiation that can be used to removethe second electron from a calcium atom?

a. 208 nm
b. 52 nm
c. 104 nm
d. 416 nm
e. 16.6 nm The second ionization energy (the energy required to removethe second outermost electron) of calcium is 11.9 eV. Determine themaximum wavelength of incident radiation that can be used to removethe second electron from a calcium atom? 208 nm 52 nm 104 nm 416 nm 16.6 nm

Explanation / Answer

The second ionization energy (the energy required to remove thesecond outermost electron) of calcium is 11.9 eV. Determine themaximum wavelength of incident radiation that can be used to removethe second electron from a calcium atom? E = h*frequency = h*c/wavelength wavelength = h*c/E = (6.626e-34 Js)*(3e8m/s)/11.9 eV *(1eV/1.60e-19J)*(1e9 nm/1m) = 104.4 nm (nanometers) h = planck's constant c = speed of light

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