Question

If the ionization potential of an atom is $20V$, its first excitation potential will be-
A.$5{\text{ V}}$
B.$10{\text{ V}}$
C.${\text{15 V}}$
D.$30{\text{ V}}$

Answer 1

Hint: Ionization potential is defined as the amount of energy required to remove an electron from an isolated atom or molecule.
Use the formula of ionization potential which is given as-
$ \Rightarrow $ E=$13.6 \times {Z^2}$ where Z is the orbit. Put the given value in the formula to obtain the value of orbit.
Then use the formula of first excitation potential which is given as-
$ \Rightarrow $ V=$ - 13.6 \times \left[ {\dfrac{{{Z^2}}}{4} - \dfrac{{{Z^2}}}{1}} \right]$
Where V is the ionization potential, Z is the orbit. Substitute the value of Z in the given formula and solve it to get the answer.

Complete step by step answer:
Given the ionization potential=$20V$
We have to find its first excitation potential.
We know that the formula of potential is given as-
$ \Rightarrow $ E=$13.6 \times {Z^2}$ where Z is the orbit.
So we know the value of ionization potential. On putting the value in the formula, we get-
$ \Rightarrow 20 = 13.6\times {{\text{Z}}^2}$
On adjusting we get-
$ \Rightarrow {Z^2} = \dfrac{{20}}{{13.6}}$ --- (i)
Now the formula of first excitation potential is given as-
$ \Rightarrow $ V=$ - 13.6 \times \left[ {\dfrac{{{Z^2}}}{4} - \dfrac{{{Z^2}}}{1}} \right]$
On putting the value of eq. (i) in the above formula, we get-
$ \Rightarrow $ V=$ - 13.6 \times \left[ {\dfrac{{\dfrac{{20}}{{13.6}}}}{4} - \dfrac{{\dfrac{{20}}{{13.6}}}}{1}} \right]$
Then we can write the above equation as-
$ \Rightarrow $ V=$ - 13.6 \times \left[ {\dfrac{{20}}{{4 \times 13.6}} - \dfrac{{20}}{{13.6}}} \right]$
Now taking $13.6$ common from the bracket we get,
$ \Rightarrow $ V=$ - \dfrac{{13.6}}{{13.6}} \times \left[ {\dfrac{{20}}{4} - 20} \right]$
On diving the terms we get,
$ \Rightarrow $ V=$ - 1 \times \left[ {5 - 20} \right]$
On subtraction, we get-
$ \Rightarrow $ V=$ - 1 \times \left[ { - 15} \right]$
Now we know that$\left( - \right) \times \left( - \right) = + $, so on applying this in the formula, we get-
$ \Rightarrow $ V=$15{\text{ V}}$
So the first excitation energy is $15{\text{V}}$.

Hence the correct answer is C.

Note:
There is a slight difference between the ionization energy and ionization potential-
1.Excitation energy is the energy required to excite an electron from its ground state to an excited state.
2.The first excitation energy is the energy required to excite an electron to the first excitation state.
3.The first excitation potential is the measure of this energy in Volts.
4. Ionization energy is the energy required to excite an atom to an infinite state. Ionization potential is the measure of this energy in volts.