If wavelength of photon is 2.2×`10 ^(−9)` m, h=6.6×`10 ^(−34)`Jsec, then momentum of photon is

Options

A. `6×10^(−23)kgms^(−1)`
B. `3×10^(−23)kgms^(−1)`
C. `3×10^(−25)kgms^(−1)`
D. `6×10^(−25)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) = 2.2 × 10^(-9) m
- Planck's constant (h) = 6.6 × 10^(-34) J·s

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

### Step 3: Rearrange the equation to solve for momentum (P)
From the de Broglie equation, we can rearrange it to find momentum:
\[ P = \frac{h}{\lambda} \]

### Step 4: Substitute the values into the equation
Now, substitute the values of h and λ into the equation:
\[ P = \frac{6.6 \times 10^{-34} \text{ J·s}}{2.2 \times 10^{-9} \text{ m}} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ P = \frac{6.6}{2.2} \times \frac{10^{-34}}{10^{-9}} \]
\[ P = 3 \times 10^{-25} \text{ kg·m/s} \]

### Final Answer
The momentum of the photon is:
\[ P = 3 \times 10^{-25} \text{ kg·m/s} \]

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