Answer:
Ionization potential of C⁺⁵ is 489.6 eV.
Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.
Explanation:
The ionization potential of hydrogen like atoms is given by the relation :
[tex]E = \frac{13.6Z^{2} }{n^{2} } eV[/tex] .....(1)
Here E is ionization potential, Z is atomic number and n is the principal quantum number which represents the state of the atom.
In this problem, the ionization potential of Carbon atom is to determine.
So, substitute 6 for Z and 1 for n in the equation (1).
[tex]E = \frac{13.6\times(6)^{2} }{1^{2} }[/tex]
E = 489.6 eV
The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :
[tex]\frac{1}{\lambda} =RZ^{2}[\frac{1}{n_{1} ^{2}}-\frac{1}{n_{2} ^{2} }][/tex] ......(2)
R is Rydberg constant, n₁ and n₂ are the transition states of the atom.
Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).
[tex]\frac{1}{\lambda} =1.09\times10^{7} \times6^{2}[\frac{1}{2 ^{2}}-\frac{1}{3 ^{2} }][/tex]
[tex]\frac{1}{\lambda}[/tex] = 5.45 x 10⁷
λ = 1.83 x 10⁻⁸ m
