The threshold frequency for the ejection of electrons from potassium metal is `5.3xx10^(14) s^(-1)`. Will the photon of a radiation having energy `3.3xx10^(-19) J` exhibit photoelectric effect? `(h=6.626xx10^(-34) Js)`

Explanation

To determine whether the photon of radiation with energy \(3.3 \times 10^{-19} \, \text{J}\) will exhibit the photoelectric effect on potassium metal, we need to compare the energy of the photon with the work function of the metal, which is related to the threshold frequency.

### Step-by-Step Solution:

**Step 1: Calculate the work function (φ) of potassium metal.**

The work function can be calculated using the formula:
\[ \phi = h \cdot f_0 \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(f_0\) is the threshold frequency (\(5.3 \times 10^{14} \, \text{s}^{-1}\)).

Substituting the values:
\[ \phi = 6.626 \times 10^{-34} \, \text{Js} \times 5.3 \times 10^{14} \, \text{s}^{-1} \]

**Step 2: Perform the multiplication.**

Calculating the above expression:
\[ \phi = 6.626 \times 5.3 \times 10^{-34} \times 10^{14} \]
\[ \phi = 35.1 \times 10^{-20} \, \text{J} = 3.51 \times 10^{-19} \, \text{J} \]

**Step 3: Compare the energy of the photon with the work function.**

The energy of the photon is given as \(E = 3.3 \times 10^{-19} \, \text{J}\).

Now we compare:
- Work function \( \phi = 3.51 \times 10^{-19} \, \text{J} \)
- Photon energy \( E = 3.3 \times 10^{-19} \, \text{J} \)

**Step 4: Determine if the photoelectric effect occurs.**

For the photoelectric effect to occur, the energy of the photon must be greater than or equal to the work function:
\[ E \geq \phi \]
In this case:
\[ 3.3 \times 10^{-19} \, \text{J} < 3.51 \times 10^{-19} \, \text{J} \]

Since the energy of the photon is less than the work function, the photoelectric effect will not occur.

### Conclusion:
The photon of radiation with energy \(3.3 \times 10^{-19} \, \text{J}\) will **not** exhibit the photoelectric effect on potassium metal.