Find the number of photons of radiation of frequancy `5xx10^(13)s^(-1)` that must be absorbed in order to melt one g ice when the latent heat of fusion of ice is `330J//s`.

Explanation

To solve the problem, we need to find the number of photons required to melt one gram of ice using the given frequency and the latent heat of fusion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between energy, number of photons, and frequency
The energy \( E \) of \( n \) photons can be expressed using the formula:
\[ E = n \cdot h \cdot \nu \]
where:
- \( E \) is the total energy required (in joules),
- \( n \) is the number of photons,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( \nu \) is the frequency of the radiation (in s\(^{-1}\)).

### Step 2: Identify the energy required to melt one gram of ice
The latent heat of fusion of ice is given as \( 330 \, \text{J/g} \). Since we are melting 1 gram of ice, the energy required \( E \) is:
\[ E = 330 \, \text{J} \]

### Step 3: Substitute the known values into the energy equation
We can rearrange the energy equation to solve for \( n \):
\[ n = \frac{E}{h \cdot \nu} \]

Substituting the values:
- \( E = 330 \, \text{J} \)
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( \nu = 5 \times 10^{13} \, \text{s}^{-1} \)

We get:
\[ n = \frac{330}{6.626 \times 10^{-34} \times 5 \times 10^{13}} \]

### Step 4: Calculate the number of photons
Now, we perform the calculation:
1. Calculate \( h \cdot \nu \):
\[ h \cdot \nu = 6.626 \times 10^{-34} \times 5 \times 10^{13} = 3.313 \times 10^{-20} \, \text{J} \]
2. Now substitute this back into the equation for \( n \):
\[ n = \frac{330}{3.313 \times 10^{-20}} \approx 9.95 \times 10^{21} \]

### Step 5: Round to significant figures
Rounding \( 9.95 \times 10^{21} \) gives approximately:
\[ n \approx 10 \times 10^{21} = 1 \times 10^{22} \]

### Final Answer
The number of photons required to melt one gram of ice is approximately:
\[ n \approx 1 \times 10^{22} \]

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