The de-Broglie wavelength of an electron, an `alpha` -particle and a proton are `lambda_(e), lambda_(alpha), lambda_(p)`. Which is wrong from the following:

Options

A. `lambda_(e) lt lambda_(p)`
B. `lambda_(p) gt lambda_(alpha)`
C. `lambda_(e) gt lambda_(alpha)`
D. `lambda_(alpha)ltlambda_(p)ltlambda_(e)`

Answer

To determine which statement is incorrect regarding the de-Broglie wavelengths of an electron, an alpha particle, and a proton, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Analyze the mass of the particles
- The mass of an electron (\( m_e \)) is very small (approximately \( 9.11 \times 10^{-31} \) kg).
- The mass of a proton (\( m_p \)) is about \( 1 \) amu (approximately \( 1.67 \times 10^{-27} \) kg).
- The mass of an alpha particle (\( m_{\alpha} \)) is about \( 4 \) amu (approximately \( 6.64 \times 10^{-27} \) kg).

### Step 3: Consider the kinetic energy
If we assume that all three particles have the same kinetic energy (\( KE \)), we can express the kinetic energy as:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can derive the velocity (\( v \)):
\[ v = \sqrt{\frac{2KE}{m}} \]

### Step 4: Relate the de-Broglie wavelength to mass
Substituting the expression for \( v \) into the de-Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{m \sqrt{\frac{2KE}{m}}} = \frac{h}{\sqrt{2mKE}} \]
This shows that the de-Broglie wavelength is inversely related to the square root of the mass when kinetic energy is constant.

### Step 5: Compare the wavelengths
Since the mass order is:
\[ m_{\alpha} > m_p > m_e \]
The order of the de-Broglie wavelengths will be the reverse:
\[ \lambda_e > \lambda_p > \lambda_{\alpha} \]
This means:
- The electron has the longest wavelength.
- The proton has a shorter wavelength than the electron but longer than the alpha particle.
- The alpha particle has the shortest wavelength.

### Step 6: Identify the incorrect statement
Based on the analysis:
- If a statement claims that the wavelength of the alpha particle is longer than that of the proton or the electron, it is incorrect.
- The correct order of wavelengths is \( \lambda_e > \lambda_p > \lambda_{\alpha} \).

### Conclusion
Thus, the incorrect statement among the options provided is the one that suggests the alpha particle has a longer wavelength than the proton or electron.