The kinetic energy of an electron in an orbit of hydrogen atom is 3.4 ev/atom. Then identify the correctly matched set for that electron
`{:(,"LIST - I",,"LIST - II",),((A),"Potential energy",(1),1.09 xx 10^(8) cm//sec,),((B),"Total energy",(2),2.11 xx 10^(-8) cm,),((C),"Velocity",(3),"-6.8 ev/atom",),((D),"Its distance from nucleus",(4),"-3.4 ev/atom",):}`
The correct match is

Options

A. A-4,B-3, C-2, D-1
B. A-3, B-4, C-1, D-2
C. A-2, B-1, C-4, D-3
D. A-3, B-4, C-2, D-1

Answer

To solve the problem, we need to analyze the kinetic energy of the electron in a hydrogen atom and relate it to potential energy, total energy, velocity, and the distance from the nucleus.

### Step-by-Step Solution:

1. **Given Information**:
- Kinetic Energy (KE) of the electron = 3.4 eV/atom.

2. **Total Energy (TE)**:
- For a hydrogen atom, the total energy (TE) is given by the formula:
\[ TE = -\frac{KE}{2} \]
- Therefore, substituting the kinetic energy:
\[ TE = -3.4 \, \text{eV} \]

3. **Potential Energy (PE)**:
- The potential energy (PE) is related to kinetic energy by the formula:
\[ PE = -2 \times KE \]
- Substituting the value of KE:
\[ PE = -2 \times 3.4 \, \text{eV} = -6.8 \, \text{eV} \]

4. **Velocity (v)**:
- The velocity of the electron in the orbit can be calculated using the formula:
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
- However, for hydrogen, we can use the known value for the velocity of the electron in the first orbit, which is approximately:
\[ v \approx 1.09 \times 10^8 \, \text{cm/s} \]

5. **Distance from Nucleus (r)**:
- The distance of the electron from the nucleus in the first orbit can be calculated using Bohr's model, where the radius of the first orbit (n=1) is approximately:
\[ r \approx 0.529 \, \text{Å} = 2.1 \times 10^{-8} \, \text{cm} \]

### Matching the Values:
Now we can match the calculated values with the options given in the question:

- **Potential Energy (PE)**: -6.8 eV/atom (matches with option C (3))
- **Total Energy (TE)**: -3.4 eV/atom (matches with option B (4))
- **Velocity (v)**: 1.09 x 10^8 cm/s (matches with option A (1))
- **Distance from Nucleus (r)**: 2.1 x 10^-8 cm (matches with option D (2))

### Final Matches:
- A (Potential Energy) → 3 (-6.8 eV/atom)
- B (Total Energy) → 4 (-3.4 eV/atom)
- C (Velocity) → 1 (1.09 x 10^8 cm/s)
- D (Distance from Nucleus) → 2 (2.1 x 10^-8 cm)

### Correct Match:
- A → 3
- B → 4
- C → 1
- D → 2