What is the minimum energy that proton must possess in order to produce photoelectric effect with platinum metal? The threshold frequency for platinum is 1.3×`10^15s^(−1)`

Options

A. 8.6×`10^12` erg
B. 8.6×`10^14` erg
C. 3.6×`10^12` erg
D. 6.6×`10^12` erg

Explanation

To find the minimum energy that a proton must possess in order to produce the photoelectric effect with platinum metal, we can use the formula that relates energy to frequency. The energy (E) of a photon is given by the equation:

\[ E = h \nu \]

Where:
- \( E \) = energy of the photon (in joules)
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \) J·s)
- \( \nu \) = frequency (in s\(^{-1}\))

Given:
- The threshold frequency (\( \nu \)) for platinum is \( 1.3 \times 10^{15} \) s\(^{-1}\).

### Step 1: Substitute the values into the energy equation

We can substitute the given threshold frequency into the energy equation:

\[ E = h \nu = (6.626 \times 10^{-34} \, \text{J·s}) \times (1.3 \times 10^{15} \, \text{s}^{-1}) \]

### Step 2: Calculate the energy

Now, we perform the multiplication:

\[ E = 6.626 \times 10^{-34} \times 1.3 \times 10^{15} \]

Calculating this gives:

\[ E = 8.6138 \times 10^{-19} \, \text{J} \]

### Step 3: Convert energy from joules to electronvolts

To convert joules to electronvolts (eV), we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E (\text{in eV}) = \frac{8.6138 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]

Calculating this gives:

\[ E \approx 5.39 \, \text{eV} \]

### Final Answer

The minimum energy that a proton must possess in order to produce the photoelectric effect with platinum is approximately \( 8.6138 \times 10^{-19} \) Joules or \( 5.39 \, \text{eV} \).

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