Which one is the wrong statement ?

Options

A. de-Broglie's wavelength is given by `lamda = (h)/(mv)` where m is the mass of the particle, `v =`groups velocity of the particle
B. The uncertainty principle is `Delta E xx Delta t ge (h)/(4pi)`
C. Half-filled and fully-filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
D. The energy of 2s orbitals is less than the energy of 2p orbital in case of hydrogen like atoms

Explanation

To determine which statement is incorrect among the given options, we will analyze each statement step by step.

### Step 1: Analyze Option 1
**Statement:** De Broglie wavelength is given by \( \lambda = \frac{h}{mv} \).

- **Explanation:** This is a correct statement. The de Broglie wavelength (\( \lambda \)) of a particle is indeed given by the formula \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

**Conclusion:** Option 1 is correct.

### Step 2: Analyze Option 2
**Statement:** The uncertainty principle is given by \( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \).

- **Explanation:** This is also a correct statement. The uncertainty principle states that the product of the uncertainties in position (\( \Delta x \)) and momentum (\( \Delta p \)) is greater than or equal to \( \frac{h}{4\pi} \). The derivation involving mass and acceleration leading to energy is valid.

**Conclusion:** Option 2 is correct.

### Step 3: Analyze Option 3
**Statement:** Half-filled and fully filled orbitals have greater stability due to greater exchange energy and symmetry.

- **Explanation:** This statement is correct. Half-filled and fully filled orbitals provide extra stability due to symmetry and exchange energy. For example, in the case of p orbitals, half-filled orbitals have one electron in each orbital, which minimizes electron-electron repulsion.

**Conclusion:** Option 3 is correct.

### Step 4: Analyze Option 4
**Statement:** The energy of the 2s orbital is less than the energy of the 2p orbital in hydrogen-like atoms.

- **Explanation:** This statement is incorrect. In hydrogen-like atoms, the energy levels depend only on the principal quantum number \( n \) and not on the azimuthal quantum number \( l \). For a given \( n \), all orbitals (s, p, d, etc.) have the same energy. Therefore, the energy of the 2s orbital is equal to the energy of the 2p orbital, not less.

**Conclusion:** Option 4 is the wrong statement.

### Final Answer
The wrong statement is **Option 4**: The energy of the 2s orbital is less than the energy of the 2p orbital in hydrogen-like atoms.

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