Let's break down your questions step by step, starting with the first one about the convex lens and the photoelectric current. When we focus sunlight onto a photosensitive plate using a convex lens, the amount of light energy that hits the plate is crucial for generating a photoelectric current, denoted as 'I'. Now, if we replace this lens with another one that has half the diameter and double the focal length, we need to analyze how these changes affect the light intensity and the resulting current.
Impact of Lens Changes on Photoelectric Current
When you change the diameter of the lens, you're altering the area through which light can pass. A lens with half the diameter has a smaller area, which means it collects less light. The area of a circle is given by the formula A = πr². If the radius is halved, the area becomes one-fourth of the original lens area.
Now, considering the focal length, a longer focal length means that the light rays converge more slowly, which can affect the intensity of light reaching the photosensitive plate. However, since the diameter is reduced, the primary factor here is the area reduction. Therefore, the overall intensity of light focused on the plate will decrease significantly.
- Original lens: Area = πr²
- New lens: Area = π(0.5r)² = π(0.25r²) = 0.25A
Since the intensity of light is directly related to the photoelectric current, the new current will be reduced to one-fourth of the original current, I. Thus, the new photoelectric current will be:
New Current = I/4
Analyzing the Helium Ion Collision
Now, let's move on to the second part of your question regarding the collision of a 100 eV electron with a stationary helium ion (He+). When this electron collides with the He+ ion, it can excite the ion to a higher energy level. The emitted photons have wavelengths of 1085 Å and 304 Å, which correspond to specific energy transitions in the helium ion.
To find the quantum number of the excited state, we can use the energy-wavelength relationship given by the equation:
E = hc/λ
Where:
- E is the energy of the photon
- h is Planck's constant (6.626 x 10-34 J·s)
- c is the speed of light (3 x 108 m/s)
- λ is the wavelength in meters
First, we convert the wavelengths from angstroms to meters:
- 1085 Å = 1.085 x 10-7 m
- 304 Å = 3.04 x 10-9 m
Now, we can calculate the energies of the emitted photons:
E1 = (6.626 x 10-34 J·s)(3 x 108 m/s) / (1.085 x 10-7 m)
E2 = (6.626 x 10-34 J·s)(3 x 108 m/s) / (3.04 x 10-9 m)
Calculating these will give you the energies corresponding to each photon. The total energy absorbed by the He+ ion must equal the energy of the incident electron (100 eV) minus the energy of the emitted photons. The quantum number of the excited state can be determined by comparing the energy levels of the helium ion, which can be found in tables or calculated based on the Rydberg formula for hydrogen-like atoms:
E = -13.6 eV (Z²/n²)
Here, Z is the atomic number (2 for helium), and n is the principal quantum number. By solving for n using the energies calculated, you can find the quantum number of the excited state.
In summary, the changes in the lens will reduce the photoelectric current to I/4, and the quantum number of the excited state can be determined through energy calculations based on the emitted photon wavelengths. This methodical approach helps in understanding the principles of optics and quantum mechanics effectively.
