Electrons of 0.5 eV energy are emitted from a metal surface when photons of wavelength 3000 Å are incident. The energy of electrons, when photons of 2000 Å are incident will be

Options

A. equal to 0.5 eV
B. higher then 0.5 eV
C. less then 0.5 eV
D. none of the above

Answer

To solve the problem, we need to apply the photoelectric effect principles and the equations associated with it. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light (photons) hits a metal surface, it can eject electrons if the energy of the photons is greater than the work function of the metal. The kinetic energy of the emitted electrons can be calculated using the equation:
\[ KE_{\text{max}} = h\nu - \phi \]
where:
- \( KE_{\text{max}} \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal.

### Step 2: Calculate the Work Function (\( \phi \))
From the problem, we know that when photons of wavelength \( 3000 \, \text{Å} \) are incident, electrons with energy \( 0.5 \, \text{eV} \) are emitted. We can use this information to find the work function.

1. Convert the wavelength from angstroms to meters:
\[ 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m} \]

2. Calculate the frequency (\( \nu \)):
\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{3 \times 10^{-7} \, \text{m}} = 10^{15} \, \text{Hz} \]

3. Calculate the energy of the incident photon:
\[ E = h\nu = (6.626 \times 10^{-34} \, \text{Js})(10^{15} \, \text{Hz}) \]
Convert \( E \) to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E \approx \frac{6.626 \times 10^{-34} \times 10^{15}}{1.6 \times 10^{-19}} \approx 2.48 \, \text{eV} \]

4. Use the photoelectric equation to find \( \phi \):
\[ KE_{\text{max}} = E - \phi \]
\[ 0.5 \, \text{eV} = 2.48 \, \text{eV} - \phi \]
\[ \phi = 2.48 \, \text{eV} - 0.5 \, \text{eV} = 1.98 \, \text{eV} \]

### Step 3: Calculate the Kinetic Energy for 2000 Å
Now we need to find the kinetic energy of the emitted electrons when photons of wavelength \( 2000 \, \text{Å} \) are incident.

1. Convert the wavelength:
\[ 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} = 2 \times 10^{-7} \, \text{m} \]

2. Calculate the frequency:
\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{2 \times 10^{-7} \, \text{m}} = 1.5 \times 10^{15} \, \text{Hz} \]

3. Calculate the energy of the incident photon:
\[ E = h\nu = (6.626 \times 10^{-34} \, \text{Js})(1.5 \times 10^{15} \, \text{Hz}) \]
Convert \( E \) to electron volts:
\[ E \approx \frac{6.626 \times 10^{-34} \times 1.5 \times 10^{15}}{1.6 \times 10^{-19}} \approx 2.48 \, \text{eV} \]

4. Use the photoelectric equation again:
\[ KE_{\text{max}} = E - \phi \]
\[ KE_{\text{max}} = 2.48 \, \text{eV} - 1.98 \, \text{eV} = 0.5 \, \text{eV} \]

### Step 4: Conclusion
The energy of the emitted electrons when photons of wavelength \( 2000 \, \text{Å} \) are incident will be greater than \( 0.5 \, \text{eV} \).