The momentum of a particle of wave length 1 Å is

Options

A. `6.625 xx 10^(-27) g.cm.s^(-1)`
B. `6.625 xx 10^(-19) g.cm.s^(-1)`
C. `6.625 xx 10^(-16) g.cm.s^(-1)`
D. `6.625 xx 10^(-23) g.cm.s^(-1)`

Answer

To find the momentum of a particle with a wavelength of 1 Å (angstrom), we can use the de Broglie wavelength equation. Here’s a step-by-step solution:

### Step 1: Understand the de Broglie Wavelength Equation
The de Broglie wavelength (\( \lambda \)) is given by the equation:
\[ \lambda = \frac{h}{p} \]
where:
- \( \lambda \) = wavelength
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \))
- \( p \) = momentum of the particle

### Step 2: Rearrange the Equation to Find Momentum
We can rearrange the equation to find the momentum (\( p \)):
\[ p = \frac{h}{\lambda} \]

### Step 3: Substitute the Given Values
Given that the wavelength \( \lambda = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \), we can substitute the values into the equation:
\[ p = \frac{6.626 \times 10^{-34} \, \text{J s}}{1 \times 10^{-10} \, \text{m}} \]

### Step 4: Calculate the Momentum
Now we perform the calculation:
\[ p = 6.626 \times 10^{-34} \div 1 \times 10^{-10} = 6.626 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Convert Units to Gram-Centimeter-Second
To convert \( p \) from kg m/s to g cm/s:
- 1 kg = 1000 g
- 1 m = 100 cm

Thus,
\[ p = 6.626 \times 10^{-24} \, \text{kg m/s} \times \frac{1000 \, \text{g}}{1 \, \text{kg}} \times \frac{100 \, \text{cm}}{1 \, \text{m}} = 6.626 \times 10^{-24} \times 1000 \times 100 \]
\[ p = 6.626 \times 10^{-24} \times 10^{5} \, \text{g cm/s} = 6.626 \times 10^{-19} \, \text{g cm/s} \]

### Final Answer
The momentum of the particle with a wavelength of 1 Å is:
\[ p = 6.626 \times 10^{-19} \, \text{g cm/s} \]

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