The speed of a photon is one hundredth of the speed light in vacuum. What is the de Broglie wavalengths. Assume that one mole of protons has a mass equal to one gram. `h = 6.626 xx 10^-27 erg sec`.

Options

A. `3.31 xx 10^-3 Å`
B. `1.33 xx 10^-3 Å`
C. `3.13 xx 10^-2 Å`
D. `1.31 xx 10^-2 Å`

Explanation

To find the de Broglie wavelength of a photon moving at one hundredth of the speed of light, we can follow these steps:

### Step 1: Determine the speed of the photon
The speed of light in vacuum (c) is approximately \(3 \times 10^8\) m/s. If the speed of the photon is one hundredth of the speed of light, we can calculate it as follows:

\[ v = \frac{c}{100} = \frac{3 \times 10^8 \text{ m/s}}{100} = 3 \times 10^6 \text{ m/s} \]

### Step 2: Calculate the mass of a single photon
We are given that one mole of photons has a mass of 1 gram. To find the mass of a single photon, we use Avogadro's number (\(N_A = 6.022 \times 10^{23}\)):

\[ m = \frac{1 \text{ gram}}{N_A} = \frac{1 \text{ g}}{6.022 \times 10^{23}} \approx 1.66 \times 10^{-24} \text{ g} \]

### Step 3: Convert mass to kg
Since we typically use SI units, we convert grams to kilograms:

\[ m = 1.66 \times 10^{-24} \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 1.66 \times 10^{-27} \text{ kg} \]

### Step 4: Use 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. Given \(h = 6.626 \times 10^{-27} \text{ erg sec}\), we need to convert it to SI units (joules):

\[ 1 \text{ erg} = 10^{-7} \text{ J} \implies h = 6.626 \times 10^{-27} \text{ erg sec} \times 10^{-7} \text{ J/erg} = 6.626 \times 10^{-34} \text{ J sec} \]

### Step 5: Substitute values into the de Broglie wavelength formula
Now we can substitute the values into the de Broglie wavelength formula:

\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J sec}}{(1.66 \times 10^{-27} \text{ kg})(3 \times 10^6 \text{ m/s})} \]

Calculating the denominator:

\[ m \cdot v = 1.66 \times 10^{-27} \text{ kg} \cdot 3 \times 10^6 \text{ m/s} = 4.98 \times 10^{-21} \text{ kg m/s} \]

Now substituting back into the equation for \(\lambda\):

\[ \lambda = \frac{6.626 \times 10^{-34}}{4.98 \times 10^{-21}} \approx 1.33 \times 10^{-13} \text{ m} \]

### Step 6: Convert to centimeters
To convert from meters to centimeters:

\[ \lambda \approx 1.33 \times 10^{-13} \text{ m} \times 100 \text{ cm/m} = 1.33 \times 10^{-11} \text{ cm} \]

### Final Result
The de Broglie wavelength of the photon is approximately:

\[ \lambda \approx 1.33 \times 10^{-11} \text{ cm} \]