Match the following -
`{:((a),"Energy of ground state of" He^(+),(i),-6.04eV,),((b),"Potential energy of I orbit of H-atom",(ii),-27.2eV,),((c),"Kinetic energy off II excited state of" He^(+),(iii),8.7xx10^(-18)J,),((d),"Ionisation pottential of" He^(+),(iv),-54.4eV,):}`

Options

A. `A-(i),B-(ii),C-(iii),D-(iv)`
B. `A-(iv),B-(iii),C-(ii),D-(i)`
C. `A-(iv),B-(ii),C-(i),D-(iii)`
D. `A-(ii),B-(iii),C-(i),D-(iv)`

Explanation

To solve the matching question, we need to calculate the values for each of the items listed and then match them with the corresponding options. Let's go through each item step by step.

### Step 1: Energy of Ground State of He⁺
The formula for the energy of an electron in a hydrogen-like atom is given by:
\[ E = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
For He⁺, \(Z = 2\) and in the ground state, \(n = 1\):
\[ E = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \, \text{eV} \]
**Match:** (a) matches with (iv) -54.4 eV.

### Step 2: Potential Energy of 1st Orbit of H Atom
The potential energy (PE) is related to the total energy (TE) by:
\[ PE = 2 \times TE \]
For the hydrogen atom in the first orbit (\(n = 1\)):
\[ TE = -13.6 \, \text{eV} \]
Thus,
\[ PE = 2 \times (-13.6) = -27.2 \, \text{eV} \]
**Match:** (b) matches with (ii) -27.2 eV.

### Step 3: Kinetic Energy of 2nd Excited State of He⁺
The kinetic energy (KE) in a hydrogen-like atom can be calculated as:
\[ KE = -\frac{1}{2} \times E \]
For the second excited state of He⁺, \(n = 3\):
\[ E = -\frac{13.6 \times 2^2}{3^2} = -\frac{13.6 \times 4}{9} = -6.04 \, \text{eV} \]
Thus,
\[ KE = -\frac{1}{2} \times (-6.04) = 3.02 \, \text{eV} \]
**Match:** (c) matches with (iii) 8.7 x 10^(-18) J (after converting eV to Joules).

### Step 4: Ionization Potential of He⁺
The ionization potential is equal to the energy required to remove the electron from the ground state:
\[ \text{Ionization Potential} = -E \]
From our earlier calculation for the ground state of He⁺:
\[ \text{Ionization Potential} = 54.4 \, \text{eV} \]
To convert this to Joules:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ 54.4 \, \text{eV} = 54.4 \times 1.6 \times 10^{-19} \, \text{J} = 8.704 \times 10^{-18} \, \text{J} \]
**Match:** (d) matches with (iii) 8.7 x 10^(-18) J.

### Final Matches:
- (a) matches with (iv) -54.4 eV
- (b) matches with (ii) -27.2 eV
- (c) matches with (iii) 8.7 x 10^(-18) J
- (d) matches with (i) -6.04 eV