Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
I'd say diffraction since sound waves can bend around objects like corners. Let's say you're in the hallway and you can hear sound coming from a door. The sound waves diffract around the door and spread out into the hallway, making it possible for you to hear.
Also, you can hear it before you see it because light waves are shorter than sound waves and hardly diffract around doors.
Answer:
6) False
7) True
8) False
9) False
10) False
11) True
12) True
13) True
14) True
Explanation:
The spacing between two energy levels in an atom shows the energy difference between them. Clearly, B has a greater value of ∆E compared to A. This implies that the wavelength emitted by B is greater than A while B will emit fewer, more energetic photons.
When atoms jump from lower to higher energy levels, photons are absorbed. The kinetic energy of the incident photon determines the frequency, wavelength and colour of light emitted by the atom.
The energy level to which an atom is excited is determined by the kinetic energy of the incident electron. As the voltage increases, the kinetic energy of the electron increases, the further the atom is from the source of free electrons, the greater the required kinetic energy of free electron. When electrons are excited to higher energy levels, they must return to ground state.
The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4
