The threshold frequency for a metal is `10^(15)` Hz. When light of wavelenght 4000 Å is made incident on it, then

Options

A. photoelectric will be emitted from it with zero speed
B. photoelectric emission will not be started by it.
C. photoelectrons will be emitted with speed `10^(5)` m/s
D. photoelectrons will be emitted with speed `10^d(3)` m/s

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Identify the given data
- Threshold frequency (ν₀) = \(10^{15}\) Hz
- Wavelength (λ) = 4000 Å (which is \(4000 \times 10^{-10}\) m)

### Step 2: Calculate the threshold energy (E₀)
The threshold energy can be calculated using the formula:
\[ E_0 = h \cdot \nu_0 \]
Where:
- \( h \) (Planck's constant) = \(6.63 \times 10^{-34}\) J·s
- \( \nu_0 \) = \(10^{15}\) Hz

Substituting the values:
\[ E_0 = 6.63 \times 10^{-34} \times 10^{15} \]
\[ E_0 = 6.63 \times 10^{-19} \text{ J} \]

### Step 3: Calculate the incident energy (E)
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( c \) (speed of light) = \(3 \times 10^8\) m/s
- \( \lambda \) = \(4000 \times 10^{-10}\) m

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{4000 \times 10^{-10}} \]
Calculating:
\[ E = \frac{1.989 \times 10^{-25}}{4 \times 10^{-7}} \]
\[ E = 4.9725 \times 10^{-19} \text{ J} \]

### Step 4: Compare the energies
- Threshold energy (E₀) = \(6.63 \times 10^{-19} \text{ J}\)
- Incident energy (E) = \(4.9725 \times 10^{-19} \text{ J}\)

### Step 5: Conclusion
Since the threshold energy (E₀) is greater than the incident energy (E), the photoelectric effect will not occur.

### Final Answer
The photoelectric effect does not take place.

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