The minimum frequency for photoelectric effect on a metal is `7xx10^(14)` Hz, Find the work function of the metal.

Answer

To find the work function of the metal given the minimum frequency for the photoelectric effect, we can follow these steps:

### Step 1: Understand the relationship between work function and frequency
The work function (Φ) of a metal is the minimum energy required to eject an electron from the surface of the metal. This energy can be calculated using the formula:
\[ Φ = h \cdot \nu_0 \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \))
- \( \nu_0 \) is the threshold frequency (given as \( 7 \times 10^{14} \, \text{Hz} \))

### Step 2: Substitute the values into the formula
Substituting the values into the equation:
\[ Φ = (6.63 \times 10^{-34} \, \text{Js}) \cdot (7 \times 10^{14} \, \text{Hz}) \]

### Step 3: Perform the multiplication
Now, we calculate the product:
\[ Φ = 6.63 \times 7 \times 10^{-34} \times 10^{14} \]
Calculating \( 6.63 \times 7 \):
\[ 6.63 \times 7 = 46.41 \]
Now, combining the powers of ten:
\[ 10^{-34} \times 10^{14} = 10^{-20} \]
Thus, we have:
\[ Φ = 46.41 \times 10^{-20} \, \text{J} \]

### Step 4: Convert to standard scientific notation
To express \( 46.41 \times 10^{-20} \) in standard scientific notation:
\[ Φ = 4.641 \times 10^{-19} \, \text{J} \]

### Final Answer
The work function of the metal is:
\[ Φ \approx 4.64 \times 10^{-19} \, \text{J} \]

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