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 follow these steps:

### Step 1: Understand the given data
- The threshold wavelength (\( \lambda_{th} \)) for the metal surface is 330 Å.
- The wavelength of the incident radiation (\( \lambda \)) is 1100 Å.

### Step 2: Calculate the threshold energy
The threshold energy (\( E_{th} \)) can be calculated using the formula:
\[ E_{th} = \frac{hc}{\lambda_{th}} \]
Where:
- \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \) eV·s)
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s)
- \( \lambda_{th} \) is in meters (1 Å = \( 10^{-10} \) m)

However, for convenience, we can use the value of \( hc \) in eV·Å, which is approximately \( 12400 \) eV·Å. Thus, we can directly substitute:
\[ E_{th} = \frac{12400 \text{ eV·Å}}{330 \text{ Å}} = 37.57 \text{ eV} \]

### Step 3: Calculate the incident energy
Next, we calculate the incident energy (\( E_{inc} \)) using the same formula:
\[ E_{inc} = \frac{hc}{\lambda} \]
Substituting the values:
\[ E_{inc} = \frac{12400 \text{ eV·Å}}{1100 \text{ Å}} = 11.27 \text{ eV} \]

### Step 4: Determine the maximum kinetic energy of emitted photoelectrons
The maximum kinetic energy (\( KE_{max} \)) of the emitted photoelectrons can be calculated using the equation:
\[ KE_{max} = E_{inc} - E_{th} \]
Substituting the values we found:
\[ KE_{max} = 11.27 \text{ eV} - 37.57 \text{ eV} = -26.30 \text{ eV} \]

### Step 5: Conclusion
Since the maximum kinetic energy is negative, it indicates that the incident energy is less than the threshold energy. Therefore, no photoelectrons will be emitted.

### Final Answer
No photoelectrons are emitted because the incident energy is less than the threshold energy.

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