Respuesta :
Answer:
(a) 13.6 eV
(b) 10.2 V
Explanation:
a) Ionization potential energy is the least requirement of energy to fetch a neutral atom from its initial state to its ionized state i.e basically the minimum energy required to excite an electron from n=1 to infinity.
Energy of a level, n, in Hydrogen atom is, [tex]E_{n}=-\frac{13.6}{n^{2} }[/tex]
Now ionization potential can be calculated as
[tex]E_{\infty } -E_{1 }[/tex]
Substitute all the value of energy and n in above equation.
[tex]=-\frac{13.6}{\infty^{2} }--\frac{13.6}{1^{2}}\\=13.6 eV[/tex]
Therefore, the ionization potential is 13.6 eV.
b) This is the energy required to excite a atom from ground state to its excited state. When electrons jumps from ground state level(n=1) to 1st excited state(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.
So, 1st excitation energy = E(n 2)- E(n = 1)
So,
[tex]=-\frac{13.6}{2^{2} }--\frac{13.6}{1^{2}}\\=-3.4eV-(-13.6eV)\\=10.2eV[/tex]
Now we can find that 1st excitation energy is 10.2 eV which gives,
[tex]eV'=10.2eV\\V'=10.2V[/tex]
Therefore, the 1st excitation potential is 10.2V.
