Consider an electron which is moving in 2nd orbit of He*. Identify the incorrect option(s) from the following with respect to the electron.
One/More correct answer(s)
A. Velocity of the moving electron is 2.18 × 10 10 cm s−1
B. Potential energy of the moving electron is -27.2 eV
C. Kinetic energy of the moving electron is -13.6 eV
D. Radius of the orbit is 1.058 Å

We are analyzing an electron in the second orbit of a hydrogen-like ion, He\(^+\) (which is singly ionized helium). In this case, the nucleus has a charge of \(Z = 2\), and the electron is in the \(n = 2\) energy level.

To determine which of the given options is incorrect, let's calculate each property step by step.

### 1. Velocity of the electron

For a hydrogen-like atom, the velocity \(v\) of an electron in orbit \(n\) is given by:
\[
v = \left(\frac{Z}{n}\right) \cdot v_0
\]
where \(v_0 = 2.18 \times 10^8 \, \text{cm/s}\) (velocity of an electron in the ground state of hydrogen).

For He\(^+\) (\(Z = 2\)) in the second orbit (\(n = 2\)):
\[
v = \left(\frac{2}{2}\right) \cdot 2.18 \times 10^8 \, \text{cm/s} = 2.18 \times 10^8 \, \text{cm/s}
\]
This means that the velocity of the electron is \(2.18 \times 10^8\) cm/s, **not \(2.18 \times 10^{10}\) cm/s**. Therefore, **Option A is incorrect**.

### 2. Potential Energy (P.E.) of the electron

The potential energy of an electron in a hydrogen-like ion is given by:
\[
\text{P.E.} = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2}
\]
For \(Z = 2\) and \(n = 2\):
\[
\text{P.E.} = -\frac{(2)^2 \cdot 13.6}{2^2} = -\frac{4 \cdot 13.6}{4} = -13.6 \, \text{eV}
\]
So, the potential energy is \(-13.6\) eV, not \(-27.2\) eV. Therefore, **Option B is incorrect**.

### 3. Kinetic Energy (K.E.) of the electron

The kinetic energy of an electron in orbit is half of the magnitude of the potential energy:
\[
\text{K.E.} = -\frac{1}{2} \cdot \text{P.E.}
\]
Since \(\text{P.E.} = -13.6 \, \text{eV}\):
\[
\text{K.E.} = -\frac{1}{2} \times (-13.6) = 13.6 \, \text{eV}
\]
Thus, the kinetic energy is \(13.6\) eV, not \(-13.6\) eV. Therefore, **Option C is incorrect**.

### 4. Radius of the orbit

The radius of an electron in the \(n\)-th orbit for a hydrogen-like ion is given by:
\[
r_n = \frac{n^2 a_0}{Z}
\]
where \(a_0 = 0.529 \, \text{Å}\) is the Bohr radius.

For \(Z = 2\) and \(n = 2\):
\[
r_2 = \frac{(2)^2 \times 0.529}{2} = \frac{4 \times 0.529}{2} = 1.058 \, \text{Å}
\]
Therefore, the radius is \(1.058\) Å, which matches the given value. Thus, **Option D is correct**.

### Final Answer:
The **incorrect options** are:
- **A**, **B**, and **C**.

---

Do you want further clarification on any of these steps or explanations? 

Here are five related questions to explore:
1. How is the energy of an electron in a hydrogen-like atom calculated?
2. What is the relationship between potential energy and kinetic energy in atomic orbitals?
3. How does the atomic radius change with different atomic numbers in hydrogen-like ions?
4. Can the Bohr model be applied to multielectron atoms?
5. Why is the electron's kinetic energy always positive?

**Tip:** In hydrogen-like atoms, the potential energy is always twice the magnitude of the kinetic energy but negative, due to the attractive electrostatic force between the nucleus and the electron.

Consider an electron which is moving in 2nd orbit of He*. Identify the incorrect option(s) from the following with respect to the electron.
One/More correct answer(s)
A. Velocity of the moving electron is 2.18 × 10^10 cm s−1
B. Potential energy of the moving electron is -27.2 eV
C. Kinetic energy of the moving electron is -13.6 eV
D. Radius of the orbit is 1.058 Å

Analyze the properties of an electron in the 2nd orbit of He*. Learn how to calculate velocity, potential energy, kinetic energy, and orbital radius, and identify the incorrect options. This problem uses the Bohr model and is suitable for undergraduate physics students.

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