The photoelectric threshold for a certain metal surface is 330 Å . What is the maximum kinetic energy of the photoelectrons emitted , if radiations of wavelength `1100 Å ` are used ?

Options

A. 1 eV
B. 2 eV
C. 7.5 eV
D. No electrons is emitted

Answer

To solve the problem, we will use the principles of the photoelectric effect and Einstein's photoelectric equation. Here are the steps to find the maximum kinetic energy of the photoelectrons emitted:

### Step 1: Understand the given data
- **Threshold wavelength (λ₀)**: 330 Å (angstrom)
- **Wavelength of incident radiation (λ)**: 1100 Å (angstrom)

### Step 2: Convert the wavelengths to meters
1 Å = \(10^{-10}\) meters
- λ₀ = 330 Å = \(330 \times 10^{-10}\) m = \(3.3 \times 10^{-8}\) m
- λ = 1100 Å = \(1100 \times 10^{-10}\) m = \(1.1 \times 10^{-7}\) m

### Step 3: Use Einstein's photoelectric equation
Einstein's photoelectric equation is given by:
\[ E = W_0 + KE_{max} \]
Where:
- \(E\) = energy of the incident photon
- \(W_0\) = work function (energy required to remove an electron)
- \(KE_{max}\) = maximum kinetic energy of the emitted photoelectron

### Step 4: Calculate the energy of the incident photon (E)
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \(h\) = Planck's constant = \(6.626 \times 10^{-34}\) J·s
- \(c\) = speed of light = \(3 \times 10^8\) m/s
- \(\lambda\) = wavelength of the incident radiation

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3 \times 10^8 \, \text{m/s})}{1.1 \times 10^{-7} \, \text{m}} \]

### Step 5: Calculate the work function (W₀)
The work function can be calculated using the threshold wavelength:
\[ W_0 = \frac{hc}{\lambda_0} \]
Substituting the values:
\[ W_0 = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3 \times 10^8 \, \text{m/s})}{3.3 \times 10^{-8} \, \text{m}} \]

### Step 6: Determine if photoelectric emission occurs
For photoelectric emission to occur, the energy of the incident photon (E) must be greater than the work function (W₀). If \(E < W_0\), no photoelectrons will be emitted.

### Step 7: Calculate the maximum kinetic energy (KE_max)
If the emission occurs, we can find the maximum kinetic energy using:
\[ KE_{max} = E - W_0 \]

### Conclusion
In this case, since the wavelength of the incident radiation (1100 Å) is greater than the threshold wavelength (330 Å), the energy of the incident photon will be less than the work function. Therefore, no photoelectrons will be emitted, and the maximum kinetic energy of the photoelectrons is 0.

### Final Answer
The maximum kinetic energy of the photoelectrons emitted is **0 J** (no emission occurs).

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