Tuesday, August 29, 2017

A material contains a radioactive isotope that disintegrates solely by the emission of α-particles at a rate of 100 s-1.






Question 1
A material contains a radioactive isotope that disintegrates solely by the emission of α-particles at a rate of 100 s-1.

Which statement about this material is correct?
A The number of atoms in the material diminishes at a rate of 100 s-1.
B The number of neutrons in the material diminishes at a rate of 100 s-1.
C The number of nucleons in the material diminishes at a rate of 400 s-1.
D The number of protons in the material diminishes at a rate of 100 s-1.





Reference: Past Exam Paper – November 2012 Paper 12 Q39





Solution 1:
Answer: C.
A nucleon is either a proton or a neutron.

An α-particle is a helium nucleus; with nucleon number 4 and proton number 2. So, each α-decay reduces the number of nucleons in the material by 4, but does not change the number of atoms.
An atom consists of protons and neutrons, that is, an atoms consists of nucleons. Thus, during emission of α-particles, the number of nucleon decreases, but not the number of atoms. [A incorrect]

This material disintegrates by emitting 100 α-particles per second.
With a decay of an alpha particle, 2 neutrons + 2 protons are emitted. So the number of neutrons and protons in the material each diminishes at a rate of 200s-1 as 100 particles are emitted per second. [B and D incorrect]
The total number of nucleons (protons + neutrons) in the material diminishes at a rate of 400s-1.

Wednesday, July 12, 2017

A cathode-ray oscilloscope (c.r.o.) is connected to an alternating voltage. The following trace is produced on the screen. ...





Question 1
A cathode-ray oscilloscope (c.r.o.) is connected to an alternating voltage. The following trace is produced on the screen.


The oscilloscope time-base setting is 0.5 ms cm-1 and the Y-plate sensitivity is 2 V cm-1.
Which statement about the alternating voltage is correct?
A The amplitude is 3.5 cm.
B The frequency is 0.5 kHz.
C The period is 1 ms.
D The wavelength is 4 cm.





Reference: Past Exam Paper – June 2014 Paper 12 Q3





Solution 1:
Answer: B.

Amplitude is the maximum voltage. It can be obtained by considering the vertical position of the trace from the central line on the screen.
Y-plate sensitivity = 2V cm-1
Amplitude = 3.5 cm = 3.5 × 2 = 7 V
[A is incorrect]

Frequency = 1 / period = 1 / T
A value of time (not ‘wavelength’) can be obtained horizontally. [D is incorrect]
Time-base setting = 0.5 ms cm-1
Period T = 4 cm = 4 × 0.5 = 2 ms       [C is incorrect]
Frequency = 1/T = 1 / (2×10-3) = 0.5 kHz

Saturday, June 24, 2017

What, to two significant figures, are the period, the frequency and the amplitude of the wave represented by the graph?






Question 1
What, to two significant figures, are the period, the frequency and the amplitude of the wave represented by the graph?


period / s        frequency / Hz           amplitude / m
A          0.0027             370                              0.0067
B          0.0031             320                              0.013
C         0.0035             290                              0.0067
D         0.0042             240                              0.013





Reference: Past Exam Paper – June 2014 Paper 12 Q22





Solution:
Answer: C.
A graph of displacement against time is given. The y-axis gives displacement in ‘m’ and the x-axis gives time in ‘ms’.

Period can be directly obtained from the time axis.
3.5 waves correspond to 12.1 ms.
3.5 T = 12.1×10-3 s
Period T = 12.1×10-3 / 3.5 = 0.0035 s

Frequency = 1 / T = 1 / 0.0035 = 290 Hz

Amplitude is the maximum displacement. This can be obtained from the y-axis.
Amplitude = 6.7 mm = 0.0067 m
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