The second ionization energy (the energy required to remove the second outermost electron) of calcium is 11.9 eV. Determine the maximum wavelength of incident radiation that can be used to remove the second electron from a calcium atom?
A)16.6 nm
B)52 nm
C)104 nm
D)208 nm
E)416 nm

Question

Quizplus · Accepted Answer

The wavelength of incident radiation can be determined using the formula:

λ = hc/E

where λ is the wavelength, h is Planck's constant, c is the speed of light, and E is the energy required to remove the electron.

For the second ionization energy of calcium:

E = 11.9 eV = 1.91 x 10^-18 J

Substituting the values:

λ = (6.626 x 10^-34 J.s x 3 x 10^8 m/s) / (1.91 x 10^-18 J)

λ = 104.1 nm

Therefore, the maximum wavelength of incident radiation that can be used to remove the second electron from a calcium atom is 104 nm. The best choice is option C.

The Second Ionization Energy (The Energy Required to Remove the Second