Question

Calculate the second excitation potential in e.v for li+2 ion

Answer 1

To calculate the second excitation potential :
We need to use the formula :
Eo (1/n1^2-1/n2^2) * z^2
Here n1=1 and n2=3 z=3 and Eo =2.18*10^-18 J. /13.6e.v.
By substituting the values
You get the answer as +108.8 E. V.

Answer 2

Answer:

The second excitation potential for $Li^{2+}$ ion measured is $108.8eV$.

Explanation:

 The second excitation potential in eV for $Li^{2+}$ ion =?

Second excitation potential:

• The potential at which an electron is jumping directly from the first shell to the third shell.

Therefore,

Second excitation potential/energy = energy in $3^{rd}$ shell - energy in $1^{st}$ shell.

• E = $E_{3} - E_{1}$

As we know,

• Excitation potential/energy = $\frac{-13.6Z^{2} }{n^{2} }$

Here,

• Z = atomic number of lithium = $3$
• n = number of shells

Now,

• E = $E_{3} - E_{1}$
• E = $\frac{-13.6Z^{2} }{n_{3} ^{2} }- \frac{-13.6Z^{2} }{n_{1} ^{2} }$
• E = ${-13.6Z^{2} }[\frac{1}{n_{3}^{2} } - \frac{1}{n_{1}^{2} }]$
• E = ${-13.6(3)^{2} }[\frac{1}{(3)^{2} } - \frac{1}{(1)^{2} }]$
• E = $-122.4[\frac{1}{9 } - \frac{1}{1 }]$
• E = $-122.4[\frac{1-9}{9} ]$
• E = $-122.4\times[\frac{-8}{9} ]$
• E = $108.8eV$

Hence, the second excitation potential for $Li^{2+}$ ion = $108.8eV$.