`C_(60)` emerging from a source at a speed (v) has a de–Broglie wavelength of 11.0 Å. The value of v (in `ms^(-1)`) is closest to [Planck's constant h = `6.626xx10^(-34)Js`]

Options

A. 0.5
B. 2.5
C. `5.0`
D. 30

Explanation

To solve the problem, we need to find the speed \( v \) of the \( C_{60} \) molecule given its de Broglie wavelength \( \lambda \) of 11.0 Å (angstroms). We will use the de Broglie wavelength formula:

\[ \lambda = \frac{h}{mv} \]

Where:
- \( \lambda \) is the de Broglie wavelength,
- \( h \) is Planck's constant,
- \( m \) is the mass of the molecule,
- \( v \) is the speed of the molecule.

### Step 1: Convert the wavelength from angstroms to meters
Given that \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):

\[ \lambda = 11.0 \, \text{Å} = 11.0 \times 10^{-10} \, \text{m} = 1.1 \times 10^{-9} \, \text{m} \]

### Step 2: Calculate the mass of \( C_{60} \)
The molecular mass of \( C_{60} \) is calculated as follows:
- The atomic mass of carbon (C) is approximately 12 amu.
- Therefore, the mass of \( C_{60} \) is:

\[ m = 60 \times 12 \, \text{amu} = 720 \, \text{amu} \]

To convert this mass to kilograms, we use the conversion factor \( 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \):

\[ m = 720 \times 1.66 \times 10^{-27} \, \text{kg} = 1.1952 \times 10^{-24} \, \text{kg} \]

### Step 3: Rearrange the de Broglie equation to solve for \( v \)
From the de Broglie equation, we can rearrange it to find \( v \):

\[ v = \frac{h}{m\lambda} \]

### Step 4: Substitute the values into the equation
Using \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( m = 1.1952 \times 10^{-24} \, \text{kg} \), and \( \lambda = 1.1 \times 10^{-9} \, \text{m} \):

\[ v = \frac{6.626 \times 10^{-34}}{(1.1952 \times 10^{-24})(1.1 \times 10^{-9})} \]

### Step 5: Calculate \( v \)
Calculating the denominator:

\[ m\lambda = (1.1952 \times 10^{-24})(1.1 \times 10^{-9}) = 1.31472 \times 10^{-33} \]

Now substituting back to find \( v \):

\[ v = \frac{6.626 \times 10^{-34}}{1.31472 \times 10^{-33}} \approx 0.504 \, \text{ms}^{-1} \]

### Step 6: Round the answer
The value of \( v \) is approximately \( 0.5 \, \text{ms}^{-1} \).

### Final Answer
The value of \( v \) (in \( \text{ms}^{-1} \)) is closest to **0.5**.

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