The work function for sodium surface is 2.0eV and that for aluminium surface is 4.2eV. The two metals are illuminated with appropriate radiations so as to cause protoemission. Then

Options

A. the threshold frequency for sodium will be less than that for aluminium
B. the threshold frequency of sodium will be more than that of aluminium
C. both sodium and aluminium will have the same threshold frequency
D. none of the above

Explanation

To solve the problem regarding the work function and threshold frequency of sodium and aluminum, we will follow these steps:

### Step 1: Understand the Work Function
The work function (W) is the minimum energy required to remove an electron from the surface of a metal. For sodium, the work function is given as 2.0 eV, and for aluminum, it is 4.2 eV.

### Step 2: Relate Work Function to Threshold Frequency
The work function is related to the threshold frequency (ν₀) by the equation:
\[ W = h \nu_0 \]
where:
- \( W \) is the work function,
- \( h \) is Planck's constant (approximately \( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( \nu_0 \) is the threshold frequency.

### Step 3: Calculate Threshold Frequencies
From the equation, we can express the threshold frequency as:
\[ \nu_0 = \frac{W}{h} \]

For sodium:
\[ \nu_{0, \text{sodium}} = \frac{2.0 \, \text{eV}}{h} \]

For aluminum:
\[ \nu_{0, \text{aluminum}} = \frac{4.2 \, \text{eV}}{h} \]

### Step 4: Compare Threshold Frequencies
Since the work function for sodium (2.0 eV) is less than that for aluminum (4.2 eV), we can conclude:
\[ \nu_{0, \text{sodium}} < \nu_{0, \text{aluminum}} \]

### Step 5: Determine the Correct Option
Based on our comparison, we find that the threshold frequency for sodium is less than that for aluminum. Thus, the correct option is:
- **Option 1**: The threshold frequency for sodium will be less than that of aluminum.

### Final Answer
The correct answer is **Option 1**.

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