Which of the alternatives gives correct match of column_i with column-II ?
`{:("","Column-I","","Column-II"),((a),"Binding energy per nucleon for" ""^(56)Fe,(i),5.5 M eV),((b),"Energy of" alpha-"particle in geiger marsden experiment",(ii),200 M eV),((c ),"Energy of photon of visible light",(iii),8.75 M eV),((d),"Energy released in fission of a uranium nucleus",(iv),2 eV):}|`

Options

A. a(i),b(iii),c(iv),d(ii)
B. a(iii),b(i),c(ii),d(iv)
C. a(iii),b(i),c(iv),d(ii)
D. a(i),b(iv),c(ii),d(iii)

Answer

To solve the problem, we need to match the items in Column-I with their corresponding values in Column-II based on the given information. Let's analyze each option step by step.

### Step-by-Step Solution:

1. **Option (a): Binding energy per nucleon for \(^{56}Fe\)**
- The binding energy per nucleon for iron-56 is known to be approximately 8.76 MeV.
- Therefore, we match option (a) with (iii) which is 8.75 MeV (close enough to 8.76 MeV).

2. **Option (b): Energy of alpha-particle in Geiger-Marsden experiment**
- The energy of the alpha particles used in the Geiger-Marsden experiment is around 5 to 6 MeV.
- Hence, we match option (b) with (i) which is 5.5 MeV.

3. **Option (c): Energy of photon of visible light**
- The energy of a photon of visible light can be calculated using the formula \(E = \frac{12413}{\lambda}\) (where \(\lambda\) is in angstroms).
- For visible light, let's take a wavelength of around 620 nm (6200 angstroms), which gives us approximately 2 eV.
- Therefore, we match option (c) with (iv) which is 2 eV.

4. **Option (d): Energy released in fission of a uranium nucleus**
- The energy released during the fission of a uranium nucleus is typically around 200 MeV.
- Thus, we match option (d) with (ii) which is 200 MeV.

### Final Matching:
- (a) → (iii) 8.75 MeV
- (b) → (i) 5.5 MeV
- (c) → (iv) 2 eV
- (d) → (ii) 200 MeV

### Summary of Matches:
- (a) → (iii)
- (b) → (i)
- (c) → (iv)
- (d) → (ii)