If wavelength of photon is 6×`10 ^(−11)` m, h=6.6×`10 ^(−34)`Jsec, then momentum of photon is

Options

A. `4×10^(−23)kgms^(−1)`
B. `1.1×10^(−23)kgms^(−1)`
C. `6×10^(−23)kgms^(−1)`
D. `3×10^(−23)kgms^(−1)`

Explanation

To find the momentum of a photon given its wavelength and Planck's constant, we can use the de Broglie relation. Here’s a step-by-step solution:

### Step 1: Write down the given values
- Wavelength of the photon (λ) = \(6 \times 10^{-11}\) m
- Planck's constant (h) = \(6.6 \times 10^{-34}\) J·s

### Step 2: Use the de Broglie relation
The de Broglie relation states that:
\[ \lambda = \frac{h}{p} \]
Where:
- λ = wavelength
- h = Planck's constant
- p = momentum of the photon

We can rearrange this equation to solve for momentum (p):
\[ p = \frac{h}{\lambda} \]

### Step 3: Substitute the values into the equation
Now, substitute the values of h and λ into the equation:
\[ p = \frac{6.6 \times 10^{-34} \text{ J·s}}{6 \times 10^{-11} \text{ m}} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ p = \frac{6.6}{6} \times \frac{10^{-34}}{10^{-11}} = 1.1 \times 10^{-23} \text{ kg·m/s} \]

### Final Answer
The momentum of the photon is:
\[ p = 1.1 \times 10^{-23} \text{ kg·m/s} \]

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