What is the mass of photon of sodium light having a wavelength of 6980A˚ ? (Given h=6.6×`10^(−27)`ergsec)

Options

A. 3.735×`10^(−29)`g
B. 3.151×`10^(−33)`g
C. 3.735×`10^(−33)`g
D. None of these

Answer

To find the mass of a photon of sodium light with a wavelength of 6980 Å (angstroms), we can use the de Broglie wavelength equation. Here’s a step-by-step solution:

### Step 1: Convert Wavelength from Angstroms to Centimeters
Given:
- Wavelength (λ) = 6980 Å

1 Å = \(1 \times 10^{-8}\) cm

So,
\[ \lambda = 6980 \, \text{Å} \times 1 \times 10^{-8} \, \text{cm/Å} = 6.98 \times 10^{-5} \, \text{cm} \]

### Step 2: Use the de Broglie Equation
The de Broglie wavelength equation is given by:
\[ \lambda = \frac{h}{mv} \]
Where:
- \(h\) = Planck's constant = \(6.6 \times 10^{-27} \, \text{erg sec}\)
- \(m\) = mass of the photon
- \(v\) = velocity of light = \(3 \times 10^{10} \, \text{cm/sec}\)

### Step 3: Rearrange the Equation to Solve for Mass
Rearranging the equation to solve for mass \(m\):
\[ m = \frac{h}{\lambda v} \]

### Step 4: Substitute the Values into the Equation
Substituting the known values:
\[ m = \frac{6.6 \times 10^{-27} \, \text{erg sec}}{(6.98 \times 10^{-5} \, \text{cm})(3 \times 10^{10} \, \text{cm/sec})} \]

### Step 5: Calculate the Mass
Calculating the denominator:
\[ \lambda v = (6.98 \times 10^{-5} \, \text{cm})(3 \times 10^{10} \, \text{cm/sec}) = 2.094 \times 10^{6} \, \text{cm}^2/\text{sec} \]

Now substituting back into the mass equation:
\[ m = \frac{6.6 \times 10^{-27}}{2.094 \times 10^{6}} \approx 3.151 \times 10^{-33} \, \text{grams} \]

### Final Answer
The mass of the photon of sodium light having a wavelength of 6980 Å is approximately:
\[ m \approx 3.151 \times 10^{-33} \, \text{grams} \]