The photoelectric threshold of a certain metal is `3000A`. If the radiation of `2000A` is incident on the metal

Options

A. electrons will be emitted
B. positrons will be emitted
C. protons will be emitted
D. electrons will no be emitted

Explanation

To solve the problem step by step, we need to determine whether the incident radiation can cause the photoelectric effect in the given metal.

### Step 1: Understand the given data
- The photoelectric threshold wavelength (λ₀) of the metal is given as 3000 Å (angstroms).
- The wavelength (λ) of the incident radiation is given as 2000 Å.

### Step 2: Convert wavelengths to meters
To work with standard SI units, we convert the wavelengths from angstroms to meters:
- 1 Å = \(10^{-10}\) m
- Therefore,
- λ₀ = 3000 Å = \(3000 \times 10^{-10}\) m = \(3.0 \times 10^{-7}\) m
- λ = 2000 Å = \(2000 \times 10^{-10}\) m = \(2.0 \times 10^{-7}\) m

### Step 3: Calculate the threshold frequency (ν₀)
Using the formula for frequency:
\[ ν₀ = \frac{C}{λ₀} \]
Where:
- C = speed of light = \(3 \times 10^8\) m/s
- λ₀ = \(3.0 \times 10^{-7}\) m

Substituting the values:
\[ ν₀ = \frac{3 \times 10^8}{3.0 \times 10^{-7}} = 1.0 \times 10^{15} \text{ Hz} \]

### Step 4: Calculate the frequency of the incident light (ν)
Using the same formula for frequency:
\[ ν = \frac{C}{λ} \]
Where:
- λ = \(2.0 \times 10^{-7}\) m

Substituting the values:
\[ ν = \frac{3 \times 10^8}{2.0 \times 10^{-7}} = 1.5 \times 10^{15} \text{ Hz} \]

### Step 5: Compare the frequencies
Now we compare the threshold frequency (ν₀) and the frequency of the incident light (ν):
- ν₀ = \(1.0 \times 10^{15}\) Hz
- ν = \(1.5 \times 10^{15}\) Hz

Since ν (frequency of the incident light) is greater than ν₀ (threshold frequency), the condition for the photoelectric effect is satisfied.

### Conclusion
Electrons will be emitted from the metal when the radiation of 2000 Å is incident on it. Therefore, the correct answer is that electrons will be emitted.

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