Hello dear FSC students, I hope you all are doing great. In today’s tutorial, we will have a look at 2nd Year Physics Chapter 20 Solved Questions. I have started a series of tutorials related to the solution of questions given in 2nd-year physics. In the previous tutorial, I have discussed all the questions of chapter20 with the detailed.
In this post, we will have a detailed look at all the questions given in chapter 21 of FSC physics. I will explain all the questions one by one and give an answer to every question in a simple way. So let’s get started with 2nd Year Physics Chapter 21 Exercise Solved Questions.
2nd Year Physics Chapter 21 Exercise Solved Questions
- So friends let’s discuss all questions one by one with the detail.
Question 21.1
-
The nuclei of similar elements that have a different mass number and same charge number are called isotopes of an element.
- The atomic number of isotopes is similar.
- Their chemical properties are similar.
- They also have the same number of protons.
- Their number of neutrons is different.
- Their physical properties are different.
- Their mass number is different.
Question 21.2
- Less weight and stable nuclei have the same number of protons and neutrons.
- While heavyweight nuclei like 238U92 Uranium consists of 92 protons and 146 neutrons.
- The 226Ra88 Radium consists of eighty-eight protons and 138 neutrons. Due to a large difference among the number of protons and neutrons heavyweight nuclei are unstable.
Question 21.3
- If the half-life of any element is one year then after passing a one year there will be half atoms will remain behind un decay.
- After further one year, half of un decay atoms will leaving behind 1/4 of atoms un-decay.
- From this, you can see that after 2 years element is not completely vanished but only 3/4 of total atoms decayed.
Question 21.4
- Suppose that at the start there are ‘N’ numbers of radioactive atoms exits. After one 1/2 life, half atoms will decay, after the second year, N/4 atoms will decay.
- After 2 years decayed atoms will be (N/2 + N/4) = 3/4th of the radioactive element.
Question 21.5
- The half-life of 226Ra88 is 1.6 * 103 so after 1.6 * 103 1/2 of 226Ra88 will decay.
- After another half-life half atoms will decay so there is infinity time is required to complete decay of an atom.
- Due to this reason still, we have 226Ra88 atoms un-decay.
Question 21.6
- There are three techniques through which electromagnetic rays interact with the matter. These techniques are given here.
- Photoelectric Effect
- Pair production
- Compton Effect
Question 21.
- As we know the charge on alpha particle is plus 2 and beta particle is a negative charge.
- Due to charges, these 2 particles have the ability to ionize the atoms without colliding the electrons.
- When the alpha particle comes close to an atom it gets electron and ionized the atom.
- When beta particle comes close to an atom it removes the electrons from that atom.
Question 21.8
- The particle whose ionizing power is large will lose all energy in a very short time interval.
- While particle with less ionize power can move to a larger distance since it will generate less number of ions.
Question 21.9
- These are some information given by the Wilson cloud chamber.
- The track for alpha particles is straight, continuous, and have large area due to the large mass and high ionizing power.
- For beta particles track is thin, short, and discontinuous due to less weight and less value of ionizing energy than the alpha particle.
- There is no specific track for gamma rays due to high penetrating power and lesser ionizing energy.
Question 21.10
- For the detection of alpha particles, Geiger Muller has thin window for the detection of alpha particles.
- Since this gives the easiest path for less penetrating alpha particles to go in the tube.
- While to detect gemma particles there is no need of any window because the penetration speed of gemma rays is high.
Question 21.11
- Its working principle based on the creation of pairs of electrons and holes after absorption of energy from the incident photons of light.
- Due to these carriers’ current flows used in detection applications.
Question 21.12
- The amount of mass of nuclear fuel that is enough to receive mostly neutrons for a self-sustained fission chain reaction.
Question 21.13
Advantage of Nuclear Power Fossil
- For the generation of electricity, this is cheap.
- These are permanent.
- Smoke does not generate through them.
- It exists in large quantities.
Advantage of Fuel Generated Fossil
- These are expensive.
- These are not permanent and their life is less.
- Smoke will generate through them.
- It does not exist in large quantities.
Disadvantage
- The radiation emitted through nuclear power are dangerous.
Question 21.14
- For fusion reaction, there is a need of high energy and temperature of million degrees that can be possible to create.
Question 21.15
Advantage:
- As there is no radioactive fossils products for a fusion reaction, so it does not hazard.
- As compared to fission reaction energy given by the fusion reaction is larger.
Question 21.16
- The radiation generated by the cosmic rays and due to the existence of radioactive materials below the crest of the earth are known as background radiation.
Question 21.17
- The alpha particles will damage more blood cells than the beta particles due to large ionizing energy.
Question 21.18
- As you know that.
- ‘D’ is absorbed dose. Its formula is.
D= Energy / Mass
E = D x Mass
- As the mass of a complete body is large than the mass of hand, so for one (mGY) more energy will be absorbed.
Question 21.19
Radioactive Tracer:
- To detect the position of an element through the mechanical system, biological and chemical systems radioisotopes are used.
- For finding the proper working of the thyroid gland radioactive iodine is used.
Question 21.20
- Radiotherapy removes the cells of cancer by distracting their DNA. For this therapy, high energy radiation is used to distract the cells.
- Gamma rays having cobalt 60 are used in the treatment of cancer.
- Radioactive iodine-131 is used to remove thyroid gland cancer.
- For the treatment of skin cancer, phosphorous 32 is used.
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