A light beam of interest `3.01xx10^(-3) W m ^(-2)` is incident normally on the square -sapped metal surface of side 1 cm and with work function 2 eV . The intensity of light is equal distributed among the three constituents' wavelength , 400 nm, 550 nm and 650 nm. Each photon emits one electron without any loss in the energy.
Choose the correct option from the following regarding the photoelectric emission.

Options

A. Electrons emitted due to light of wavelength 400 nm only
B. Electrons emitted due to light of wavelength 650 nm only
C. Electrons emitted due to light of wavelengths 440 nm and 550 nm only
D. Electrons emitted due to light of wavelengths 550 nm and 650 nm only

Explanation

To solve the problem step by step, we need to determine which wavelengths of light can cause photoelectric emission from the metal surface given its work function.

### Step 1: Understand the Work Function
The work function (W) is the minimum energy required to remove an electron from the surface of the metal. In this case, the work function is given as 2 eV.

### Step 2: Convert Work Function to Joules
To work with SI units, we need to convert the work function from electron volts to joules. The conversion factor is:
1 eV = \(1.6 \times 10^{-19}\) Joules.

So,
\[ W = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the Threshold Wavelength
The threshold wavelength (\(\lambda_0\)) can be calculated using the formula:
\[ \lambda_0 = \frac{hc}{W} \]
where:
- \(h\) (Planck's constant) = \(6.63 \times 10^{-34} \, \text{J s}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)

Substituting the values:
\[ \lambda_0 = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3.2 \times 10^{-19} \, \text{J}} \]
Calculating this gives:
\[ \lambda_0 = \frac{1.989 \times 10^{-25}}{3.2 \times 10^{-19}} = 6.19 \times 10^{-7} \, \text{m} = 619 \, \text{nm} \]

### Step 4: Compare Given Wavelengths with Threshold Wavelength
The wavelengths given in the problem are:
- 400 nm
- 550 nm
- 650 nm

We need to determine which of these wavelengths are less than or equal to the threshold wavelength (619 nm):
- 400 nm < 619 nm (can emit electrons)
- 550 nm < 619 nm (can emit electrons)
- 650 nm > 619 nm (cannot emit electrons)

### Step 5: Conclusion
Electrons will be emitted due to the light of wavelengths 400 nm and 550 nm only. Therefore, the correct option is that electrons are emitted due to the light of 400 nm and 550 nm only.

### Final Answer
Electrons emitted due to the light of a wavelength of 400 nm and 550 nm only.

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