Choose the correct alternatives

A. Ratio of de-Broglie wavelengths of proton and α-particle of same kinetic energy is 2L1

B. if neutron, α-particle and β-particle all are moving with same kinetic energy , β-particle has maximum de-Broglie wavelength

C. de-Broglie hypothesis treats particles as waves

D. de-Broglie hypothesis treats waves as made of particles

Options

A. Ratio of de-Broglie wavelengths of proton and α-particle of same kinetic energy is 2L1
B. if neutron, α-particle and β-particle all are moving with same kinetic energy , β-particle has maximum de-Broglie wavelength
C. de-Broglie hypothesis treats particles as waves
D. de-Broglie hypothesis treats waves as made of particles

Answer

To solve the question, we will analyze each statement one by one and determine whether they are correct or not based on the principles of de-Broglie wavelength and kinetic energy.

### Step 1: Analyze Statement A
**Statement A:** Ratio of de-Broglie wavelengths of proton and α-particle of the same kinetic energy is 2:1.

**Solution:**
The de-Broglie wavelength (λ) is 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.

For kinetic energy (KE):
\[ KE = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2KE}{m}} \]

Substituting \( v \) into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m\sqrt{\frac{2KE}{m}}} = \frac{h}{\sqrt{2mKE}} \]

Thus, the de-Broglie wavelength is inversely proportional to the square root of the mass.

Let \( m_p \) be the mass of the proton and \( m_{\alpha} \) be the mass of the α-particle (approximately 4 times the mass of the proton):
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{m_{\alpha}}}{\sqrt{m_p}} = \sqrt{\frac{4m_p}{m_p}} = 2 \]

So, the ratio of the de-Broglie wavelengths of the proton to the α-particle is indeed 2:1.

**Conclusion:** Statement A is correct.

### Step 2: Analyze Statement B
**Statement B:** If neutron, α-particle, and β-particle all are moving with the same kinetic energy, β-particle has maximum de-Broglie wavelength.

**Solution:**
We already know that the de-Broglie wavelength is inversely proportional to the square root of the mass.

- Mass of β-particle (electron) \( m_{\beta} \approx 9.1 \times 10^{-31} \) kg
- Mass of neutron \( m_n \approx 1.67 \times 10^{-27} \) kg
- Mass of α-particle \( m_{\alpha} \approx 4 \times m_p \approx 6.64 \times 10^{-27} \) kg

Since the β-particle has the smallest mass, it will have the largest de-Broglie wavelength when all particles have the same kinetic energy.

**Conclusion:** Statement B is correct.

### Step 3: Analyze Statement C
**Statement C:** de-Broglie hypothesis treats particles as waves.

**Solution:**
The de-Broglie hypothesis indeed states that particles can exhibit wave-like behavior, which is a fundamental concept in quantum mechanics.

**Conclusion:** Statement C is correct.

### Step 4: Analyze Statement D
**Statement D:** de-Broglie hypothesis treats waves as made of particles.

**Solution:**
This statement is not correct. The de-Broglie hypothesis primarily discusses the wave nature of particles, not the particle nature of waves.

**Conclusion:** Statement D is incorrect.

### Final Answer:
The correct alternatives are A, B, and C.

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