What is the mass of photon of sodium light having a wavelength of 4880A˚ ? (Given h=6.6×`10^(−27)`ergsec)

Options

A. 3.735×`10^(−29)`g
B. 4.508×`10^(−33)`g
C. 2.735×`10^(−33`g
D. None of these

Answer

To find the mass of a photon of sodium light with a wavelength of 4880 Å (angstroms), we can use the relationship derived from the de Broglie equation. Here’s a step-by-step solution:

### Step 1: Convert Wavelength from Angstroms to Centimeters
The wavelength given is 4880 Å. We know that:
1 Å = \(10^{-8}\) cm

So, to convert 4880 Å to centimeters:
\[ \text{Wavelength} (\lambda) = 4880 \, \text{Å} \times 10^{-8} \, \text{cm/Å} = 4.88 \times 10^{-5} \, \text{cm} \]

### Step 2: Use the Photon Mass Formula
The mass of a photon can be calculated using the formula:
\[ m = \frac{h}{\lambda \cdot v} \]
where:
- \(m\) = mass of the photon
- \(h\) = Planck's constant = \(6.6 \times 10^{-27} \, \text{erg sec}\)
- \(\lambda\) = wavelength in cm = \(4.88 \times 10^{-5} \, \text{cm}\)
- \(v\) = speed of light = \(3 \times 10^{10} \, \text{cm/s}\)

### Step 3: Substitute the Values into the Formula
Now, we can substitute the values into the equation:
\[ m = \frac{6.6 \times 10^{-27} \, \text{erg sec}}{(4.88 \times 10^{-5} \, \text{cm}) \cdot (3 \times 10^{10} \, \text{cm/s})} \]

### Step 4: Calculate the Denominator
First, calculate the denominator:
\[ \lambda \cdot v = (4.88 \times 10^{-5} \, \text{cm}) \cdot (3 \times 10^{10} \, \text{cm/s}) = 1.464 \times 10^{6} \, \text{cm}^2/\text{s} \]

### Step 5: Calculate the Mass
Now substitute the denominator back into the mass equation:
\[ m = \frac{6.6 \times 10^{-27}}{1.464 \times 10^{6}} \approx 4.508 \times 10^{-33} \, \text{grams} \]

### Final Answer
The mass of the photon of sodium light having a wavelength of 4880 Å is approximately:
\[ m \approx 4.508 \times 10^{-33} \, \text{grams} \]

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