Question 23M.3.HL.TZ2.7
| Date | May 2023 | Marks available | [Maximum mark: 5] | Reference code | 23M.3.HL.TZ2.7 |
| Level | HL | Paper | 3 | Time zone | TZ2 |
| Command term | Deduce, Explain | Question number | 7 | Adapted from | N/A |
A rocket ship is at rest on the surface of a non-rotating planet X. The rocket ship contains a chamber of height 20 m. Photons are emitted with frequency 3.2 × 1010 Hz and travel from the floor of the chamber to the ceiling of the chamber. A receiver on the ceiling detects the frequency of the photons.
Explain why the frequency of the photons detected at the ceiling is less than the frequency of those emitted from the floor.
[1]
some photon energy is used to do work against the gravitational field
OR
mention of gravitational redshift ✓
OWTTE
The change in frequency detected at the ceiling as compared to the floor was measured to be 1.2 × 10−4 Hz. Deduce the gravitational field strength of planet X.
[2]
✓
= 16.9 «m s2» ✓
Award [2] for BCA
The rocket ship is launched and accelerates vertically. Explain, with reference to the equivalence principle, why the magnitude of the frequency change observed in photons emitted from floor to ceiling of the rocket ship will increase as it launches.
[2]
gravitational effects cannot be distinguished from inertial effects ✓
«apparent» acceleration within rocket is greater than g
OR
there is a stronger gravitational field
OR
the receiver is moving further away ✓
OWTTE