Dual Nature Of Radiation And Matter

The maximum kinetic energy of the photoelectrons is given by the equation: where is the energy of the photon, and is the work function. For the two photons, the maximum kinetic energies are: The speed of the photoelectron is given by: Thus, the ratio of speeds is:

Step 1: Kinetic Energy from de Broglie Wavelength:
The de Broglie wavelength of an electron is given by , where is momentum. So, momentum . The Kinetic Energy ( ) of the electron is:
Step 2: X-ray Cut-off Wavelength:
The entire kinetic energy of the electron is converted into a single photon for the minimum wavelength ( ) of X-rays produced (Duane-Hunt Law).
Step 3: Equating and Solving:
Solving for :
Step 4: Final Answer:
The minimum wavelength is .

Step 1: State Einstein's Photoelectric Equation.
The maximum kinetic energy ( ) of the emitted photoelectron is related to the energy of the incident photon ( ) and the work function ( ) of the metal by:

Step 2: Relate kinetic energy to the stopping potential.
The maximum kinetic energy is also related to the stopping potential ( ) by the equation: where is the elementary charge.

Step 3: Combine the equations and find the relation between and .
Substituting into the photoelectric equation: Solving for the stopping potential :

Step 4: Determine the slope of the versus graph.
The graph is plotted with stopping potential ( ) on the y-axis and frequency ( ) on the x-axis. Comparing the equation from Step 3 to the equation of a straight line, , the slope ( ) of the graph is:

Step 5: Calculate the numerical value of the slope.
Given Planck's constant . The elementary charge is .

The stopping potential ( ) is the potential difference required to stop the most energetic photoelectrons.
The relationship between the stopping potential and the maximum kinetic energy ( ) of the photoelectrons is:
.
Where is the elementary charge.
The maximum kinetic energy can also be calculated from the maximum velocity ( ) of the photoelectrons:
.
Where is the mass of the electron.
We are given:
m/s.
kg.
C.
First, calculate the maximum kinetic energy in Joules:
.
J.
Now, we can find the stopping potential from :
.
V.
The work function (1.5 eV) is extra information not needed to find the stopping potential from the given velocity.

Step 1: Understanding the Concept:
The production of X-rays in an X-ray tube involves accelerating electrons through a high potential difference and making them strike a metal target. The minimum wavelength (or maximum frequency) of the produced X-rays corresponds to the scenario where an electron loses all its kinetic energy in a single collision, converting it into a single X-ray photon. This is related to the cutoff wavelength of the continuous X-ray spectrum.
Step 2: Key Formula or Approach:
The kinetic energy ( ) gained by an electron accelerated through a potential difference is .
The energy ( ) of a photon of wavelength is .
For the minimum wavelength ( ), the maximum possible photon energy is equal to the electron's kinetic energy:
Rearranging for :
A very useful approximation for calculations is:
where V is the potential difference in Volts.
Step 3: Detailed Explanation:
Given values:
- Accelerating potential, .
Using the shortcut formula:
Let's check the answer in Angstroms (Å) as well, since some options are in Å. (1 nm = 10 Å).
Comparing our result with the options:
- (A) 0.248 Å
- (B) 0.41 Å
- (C) 0.099 nm
- (D) 0.062 nm
Our calculated value matches option (D).
Step 4: Final Answer:
The minimum wavelength of the X-rays produced is 0.062 nm. Therefore, option (D) is correct.

1. The de Broglie wavelength for a particle is , where is Planck's constant and is momentum.
2. For an electron accelerated through potential , kinetic energy , so .
3. Thus, . The numerical formula in Ångströms is , where is in volts.
4. Given V, .
5. Calculate Å.
6. Therefore, the correct option is (1) 0.25 Å.

• de Broglie wavelength: , .
• For proton: , alpha: .
• Equate wavelengths: Wait, alpha particle has charge 2e → .
• Hence .