Structure Of The Atom - Olympiad Level MCQ, Class 9 Science - Class 9 MCQ

Metals and hydrogen are always electropositive because:-*Hydrogen-It has 1 outermost electron and its valency is 1 so it can only gain electron but cannot lose it .So, it is placed in the first group.Hence,it is electropositive.*Metals-It has tendency to lose their electron and form positively charged ion .Hence,Metals are electropositive.

The Mg²⁺ and F^- ions differ in the following fundamental particles:
1. Electrons:
- Mg²⁺ ion has lost two electrons compared to the neutral Mg atom. It has 10 electrons.
- F^- ion has gained one electron compared to the neutral F atom. It has 10 electrons.
2. Protons:
- Both Mg²⁺ and F^- ions have the same number of protons as their respective neutral atoms.
- Mg atom has 12 protons, so Mg²⁺ ion also has 12 protons.
- F atom has 9 protons, so F^- ion also has 9 protons.
3. Neutrons:
- Both Mg²⁺ and F^- ions have the same number of neutrons as their respective neutral atoms.
- Mg atom has 12 neutrons, so Mg²⁺ ion also has 12 neutrons.
- F atom has 10 neutrons, so F^- ion also has 10 neutrons.
In conclusion:
The Mg²⁺ and F^- ions differ only in the number of electrons they possess. They have the same number of protons and neutrons.

The isotope used to remove brain tumors and treat cancer is Cobalt-60 (Co-60). Here is a detailed explanation:
Isotope Used for Brain Tumor Removal and Cancer Treatment:


Cobalt-60 (Co-60) is the isotope used for removing brain tumors and treating cancer. It is a radioactive isotope of the element cobalt.
Properties of Cobalt-60:


Cobalt-60 has the following properties that make it suitable for medical applications:

• Radioactive: Cobalt-60 is a radioactive isotope, which means it emits radiation.


• High Energy Radiation: It emits gamma rays, which are high-energy radiation that can penetrate tissues deeply.


• Long Half-Life: Cobalt-60 has a half-life of approximately 5.27 years, which allows for long-term use in medical treatments.

Medical Applications:


Cobalt-60 is commonly used in radiation therapy for cancer treatment, including the removal of brain tumors. It is used in a procedure called stereotactic radiosurgery, where a precise and concentrated dose of radiation is delivered to the tumor.

• Brain Tumor Removal: Cobalt-60 is used to target and destroy brain tumors, either by shrinking them or completely eliminating them.


• Cancer Treatment: Cobalt-60 is also used to treat various types of cancer throughout the body, including lung, prostate, and breast cancer.

Advantages of Cobalt-60:


Cobalt-60 offers several advantages in medical applications:

• High Penetrating Power: The gamma rays emitted by cobalt-60 can penetrate deep into tissues, making it effective for treating tumors located deep within the body.


• Precise Targeting: Cobalt-60 can be precisely targeted to the tumor site, minimizing damage to surrounding healthy tissues.


• Long-Lasting: The long half-life of cobalt-60 allows for extended use in medical treatments without frequent replacement.

In conclusion, Cobalt-60 is the isotope used for removing brain tumors and treating cancer. Its properties and advantages make it an effective tool in radiation therapy for cancer treatment.

Let the valency of Sn =x,

 

For SnCl₂ :

 

Since valency of Cl is -1.

 

x +(-1 × 2) = 0,

 

So,  x = 2,

 

For SnCl₄ :

 

x +(-1 × 4) = 0,

 

So, x = 4

Valency of Sulphur in SO₂ and SO₃
The valency of an element indicates the number of bonds it can form with other elements. In the case of Sulphur, its valency can vary depending on the compound it is present in. Let's determine the valency of Sulphur in SO₂ and SO₃:
1. Valency of Sulphur in SO₂
- SO₂ is a molecule where one Sulphur atom is bonded to two Oxygen atoms.
- Oxygen has a valency of 2, meaning it can form two bonds.
- By subtracting the valency of Oxygen from the total number of available electrons for Sulphur, we can determine its valency.
- Sulphur has 6 valence electrons.
- Oxygen contributes 2 electrons per bond.
- Therefore, Sulphur forms two bonds with Oxygen, resulting in a valency of 2.
2. Valency of Sulphur in SO₃
- SO₃ is a molecule where one Sulphur atom is bonded to three Oxygen atoms.
- Using the same approach as before, we can determine the valency of Sulphur in SO₃.
- Oxygen has a valency of 2, and there are three Oxygen atoms, contributing a total of 6 electrons.
- Sulphur has 6 valence electrons.
- Oxygen contributes 2 electrons per bond.
- Therefore, Sulphur forms three bonds with Oxygen, resulting in a valency of 6.
Summary:
- The valency of Sulphur in SO₂ is 2.
- The valency of Sulphur in SO₃ is 6.
Therefore, the correct answer is option D: 4, 6.

Alpha particle. Alpha particle, positively charged particle, identical to the nucleus of the helium-4 atom, spontaneously emitted by some radioactive substances, consisting of two protons and two neutrons bound together, thus having a mass of four units and a positive charge of two.

Explanation:
To determine which electronic configuration is wrong, we need to compare the given configurations with the correct electronic configurations for the respective elements.
The correct electronic configurations for the elements in the given options are as follows:
A: Be (4) = 2, 2
B: O (8) = 2, 6
C: S (16) = 2, 8, 6
D: P (15) = 2, 8, 5
Comparing these with the given options, we can see that option B is incorrect. The correct electronic configuration for oxygen (O) is 2, 6, but the given option states it as 2, 8.
Therefore, the correct answer is Option B: O (8) = 2, 8.

The mass of a proton can be determined by referring to the current accepted value in scientific literature.
Key Point: The mass of a proton is a fundamental constant in physics and is expressed in kilograms (kg) or atomic mass units (u).
To find the correct answer, we need to convert the given options to kg or u and compare them to the accepted value.
Key Point: The accepted value for the mass of a proton is approximately 1.67262192 × 10^-27 kg or 1.007276 u.
Now let's examine each option and convert them to the appropriate units:
A: 1.609 g
- To convert grams to kilograms, we divide by 1000: 1.609 g = 0.001609 kg
- This value is not close to the accepted value, so option A is incorrect.
B: 1.6 × 10^24 g
- To convert grams to kilograms, we divide by 10^3: 1.6 × 10^24 g = 1.6 × 10^21 kg
- This value is much larger than the accepted value, so option B is incorrect.
C: 1.6 × 10^-23 g
- To convert grams to kilograms, we divide by 10^-3: 1.6 × 10^-23 g = 1.6 × 10^-26 kg
- This value is close to the accepted value, so option C is a possible answer.
D: 1.6 × 10^-24 g
- To convert grams to kilograms, we divide by 10^-3: 1.6 × 10^-24 g = 1.6 × 10^-27 kg
- This value is the closest to the accepted value, so option D is the correct answer.
Therefore, the correct answer is option D: 1.6 × 10^-24 g.

Electronic configuration of noble gas atom can be identified by full Valence Electrons .
According to that of option C is correct of configuration 2,8.

Explanation:
The e/m ratio, also known as the specific charge, is the charge-to-mass ratio of a particle. It is defined as the ratio of the charge (e) to the mass (m) of the particle. The value of e/m depends on the type of particle being considered.
Let's examine each option to determine whether the e/m ratio is constant or not:
A: Cathode rays
- Cathode rays are streams of electrons that are emitted from the cathode in a vacuum tube.
- The e/m ratio for cathode rays was determined by J.J. Thomson through his experiments with cathode ray tubes.
- Thomson found that the e/m ratio for cathode rays is constant and independent of the gas used in the tube or the material of the electrodes.
- Therefore, the e/m ratio for cathode rays is constant.
B: Positive rays
- Positive rays, also known as canal rays, are streams of positively charged ions that are produced in a gas discharge tube.
- The e/m ratio for positive rays was also determined by J.J. Thomson.
- Thomson found that the e/m ratio for positive rays depends on the specific gas used in the tube and the charge of the ion.
- Therefore, the e/m ratio for positive rays is not constant.
C: α -rays
- α-rays are streams of alpha particles, which are helium nuclei consisting of two protons and two neutrons.
- The e/m ratio for α-rays is constant.
- This is because the charge (e) and mass (m) of an alpha particle are known and do not change.
- Therefore, the e/m ratio for α-rays is constant.
D: β -rays
- β-rays are streams of beta particles, which are high-energy electrons or positrons.
- The e/m ratio for β-rays is not constant.
- This is because the charge (e) and mass (m) of a beta particle can vary depending on whether it is an electron or a positron.
- Therefore, the e/m ratio for β-rays is not constant.
Conclusion:
From the given options, the e/m ratio is constant for cathode rays (option A) and α-rays (option C), but not for positive rays (option B) and β-rays (option D). Therefore, the correct answer is option C.

Size of the nucleus:
The size of the nucleus refers to the diameter or radius of the nucleus, which is the central part of an atom. The nucleus is composed of protons and neutrons, which are collectively referred to as nucleons. The size of the nucleus can be determined by measuring the nuclear radius.
Options:
A:

10^-15 cm

B:

10^-13 cm

C:

10^-10 cm

D:

10^-8 cm

Answer: B. 10^-13 cm
Explanation:
The correct answer is B, 10^-13 cm. Here's why:
- The size of the nucleus is extremely small compared to the overall size of the atom.
- The actual size of the nucleus can vary depending on the element or isotope. However, on average, the nuclear radius is approximately 10^-13 cm.
- This means that the nucleus is about 10,000 times smaller than the entire atom.
- The nucleus is so small that if an atom were the size of a football field, the nucleus would be about the size of a pea in the center.
- Option B, 10^-13 cm, is the closest approximation to the average size of the nucleus.
In summary, the size of the nucleus is approximately 10^-13 cm, which is option B.

A Danish physicist named Neil Bohr in 1913 proposed the Bohr atomic model. He modified the problems and limitations associated with Rutherford's model of an atom. Earlier in Rutherford Model, Rutherford explained in an atom a nucleus is positively charged and is surrounded by electrons (negatively charged particles).

Explanation:
Cathode rays are streams of electrons that are emitted from the cathode (negative electrode) of a high-voltage electrical discharge tube. These rays have both mass and charge, and their properties were extensively studied by scientists in the late 19th and early 20th centuries.
Properties of cathode rays:
- Mass: Cathode rays have mass. This was demonstrated by the German physicist J.J. Thomson in his experiments with cathode rays. He found that cathode rays could be deflected by electric and magnetic fields, indicating that they have mass.
- Charge: Cathode rays are negatively charged. This was also shown by Thomson in his experiments. He observed that cathode rays were deflected towards a positively charged plate, indicating that they carry a negative charge.
Conclusion:
Based on these experimental observations, it can be concluded that cathode rays have both mass and charge. Therefore, the correct answer is option D: Both mass and charge.

To determine which of the following statements is false, let's analyze each statement individually:
A: Neutron has highest mass among fundamental particles
- The neutron is one of the three fundamental particles that make up an atom, along with protons and electrons.
- The mass of a neutron is approximately equal to the mass of a proton, which is approximately 1 atomic mass unit (AMU).
- However, the mass of an electron is significantly smaller, approximately 1/1836th of the mass of a proton or neutron.
- Therefore, statement A is true, and it is not the false statement.
B: The mass of an electron is negligible
- As mentioned earlier, the mass of an electron is much smaller than the mass of a proton or neutron.
- However, in certain contexts, such as atomic and subatomic physics, the mass of an electron is considered significant.
- For example, the mass of an electron is crucial in determining the overall mass and stability of an atom.
- Therefore, statement B is true, and it is not the false statement.
C: e/m is highest for a proton
- The ratio of charge to mass (e/m) is known as the specific charge.
- The specific charge is highest for particles with the smallest mass and highest charge.
- While the proton has a positive charge and a relatively high mass, it is not correct to say that its specific charge is the highest.
- The specific charge of an electron is significantly higher than that of a proton since the electron has a much smaller mass.
- Therefore, statement C is false, and it is the false statement.
D: Charge of neutron is zero
- The neutron is an electrically neutral particle, meaning it has no net charge.
- It contains no charge and is composed of neutral quarks.
- Therefore, statement D is true, and it is not the false statement.
In conclusion, the false statement among the given options is C: e/m is highest for a proton.

The specific charge of a particle is defined as the charge per unit mass of the particle. In this case, we are comparing the specific charge of an α-proton (also known as an alpha particle) to that of another particle.
The specific charge of an α-proton can be calculated using the formula:
Specific charge = Charge / Mass
To compare the specific charge of an α-proton to that of another particle, we need to find the ratio of their charges and masses.
The charge of an α-proton is 2e, where e is the elementary charge.
The mass of an α-proton is 4 times the mass of a hydrogen atom, which is equal to 4u, where u is the atomic mass unit.
Therefore, the specific charge of an α-proton is:
Specific charge of α-proton = (2e) / (4u)
Now, let's consider another particle. Let's assume its charge is q and its mass is m.
The specific charge of this particle is:
Specific charge of particle = q/m
To find the ratio of the specific charges, we can divide the specific charge of the α-proton by the specific charge of the other particle:
Ratio of specific charges = (2e) / (4u) / (q/m)
= (2e/m) / (4u/q)
= (2e/m) * (q/4u)
= (2e * q) / (4u * m)
= (e * q) / (2u * m)
Since the α-proton and the particle have the same charge (e), the ratio of their specific charges is determined by the ratio of their masses:
Ratio of specific charges = (1) / (2u * m)
From this expression, we can see that the ratio of specific charges is inversely proportional to the mass of the particle.
Therefore, the correct answer is C: 1 : 4.

In particle physics, mesons are hadronic subatomic particles composed of one quark and one ... However, such heavy mesons are regularly created in particle ...
Spin: 0, 1
Electric charge: −1 e, 0 e, +1 e
Mass: From 134.9 MeV/c2 (; π0;); to 9.460 GeV/c2 (; ϒ;)
Discovered: 1947

To determine the nature of the species 'X' with 9 protons, 10 electrons, and 11 neutrons, we need to analyze the charge distribution.
1. Determine the charge of the protons:
- Protons have a positive charge of +1.
2. Determine the charge of the electrons:
- Electrons have a negative charge of -1.
3. Determine the total charge:
- The total positive charge from the protons is +9.
- The total negative charge from the electrons is -10.
4. Analyze the charge distribution:
- If the total charge is zero, the species is a neutral atom (A).
- If the total charge is positive, the species is a cation (C).
- If the total charge is negative, the species is an anion (D).
5. Calculate the total charge:
- The total charge is +9 - 10 = -1.
6. Determine the nature of the species:
- Since the total charge is -1, the species 'X' is an anion (D).
Therefore, the correct answer is D. An anion.

Bohr's atomic theory says that each shell has its own energy and it neither loses nor gains energy and due to this atoms are static and therefore this energy shells are called stationary states

Energy of an orbit:
The energy of an electron in an orbit around the nucleus of an atom depends on the distance from the nucleus. This is explained by the concept of energy levels or shells.
1. Energy levels or shells:
- Electrons in an atom occupy specific energy levels or shells.
- The energy levels are labeled with numbers (n=1, 2, 3, etc.), where the lowest energy level is closest to the nucleus.
- Each energy level can hold a certain maximum number of electrons.
2. Energy of an orbit:
- The energy of an electron in an orbit is quantized, meaning it can only exist at specific energy levels and not in between.
- The energy of an electron in a particular energy level is lower than the energy of an electron in a higher energy level.
- The energy of an electron in an orbit is negative.
- The energy of an electron in the first energy level (n=1) is the most negative, and it becomes less negative as we move to higher energy levels.
3. Relationship with distance from the nucleus:
- As we move away from the nucleus, the distance between the electron and the nucleus increases.
- The energy of an electron in an orbit is inversely proportional to the distance from the nucleus.
- Therefore, as we move away from the nucleus, the energy of the electron increases.
4. Answer: D:
- The correct answer is D: None of the above.
- The energy of an orbit increases as we move away from the nucleus, not decreases as option C suggests.
- The energy of an orbit does not remain the same as we move away from the nucleus, as option B suggests.
- Therefore, none of the given options accurately describe the energy of an orbit.

According to Bohr's model of the hydrogen atom, the linear velocity of the electron is quantized. Here is a detailed explanation of why this is the case:
Bohr's Model of the Hydrogen Atom
- Bohr's model was proposed by Niels Bohr in 1913 to explain the behavior of electrons in the hydrogen atom.
- According to this model, electrons orbit the nucleus in circular paths called orbits or energy levels.
- Each energy level corresponds to a specific amount of energy that the electron possesses.
Quantization of Linear Velocity
- In Bohr's model, the electron is restricted to specific energy levels or orbits.
- The electron can only occupy certain energy levels, which means it can only have certain amounts of energy.
- The energy of an electron in a specific energy level is given by the equation E = (-13.6 eV)/n^2, where E is the energy, n is the principal quantum number, and -13.6 eV is the ionization energy of hydrogen.
- The linear velocity of an electron in an orbit is directly related to its energy.
- Since the energy levels are quantized, the linear velocity of the electron is also quantized.
Implications of Quantization of Linear Velocity
- The quantization of linear velocity means that the electron can only have certain velocities while orbiting the nucleus.
- The electron cannot have any arbitrary velocity but only velocities corresponding to the allowed energy levels.
- This implies that the electron's motion in the hydrogen atom is not continuous but occurs in discrete steps or jumps between energy levels.
In conclusion, according to Bohr's model of the hydrogen atom, the linear velocity of the electron is quantized, meaning that it can only have certain velocities corresponding to the allowed energy levels. This quantization of linear velocity is a fundamental aspect of the model and helps explain the behavior of electrons in the hydrogen atom.

Indigenous Freshwater Fishes:
- Catla and Rohu are examples of indigenous freshwater fishes.
- Indigenous freshwater fishes are native to a particular region or country.
- They are usually found in lakes, rivers, and ponds within their natural habitat.
- These fishes have adapted to the specific environmental conditions of their native waters.
- Catla and Rohu are commonly found in the rivers and lakes of the Indian subcontinent.
- They are highly valued for their taste and are popular food fishes in South Asia.
- These fishes are an important part of the local freshwater ecosystems and play a role in maintaining the balance of aquatic life.
- Catla and Rohu have been traditionally farmed and consumed in India for centuries.
Exotic Freshwater Fishes:
- Exotic freshwater fishes refer to species that are not native to a particular region.
- They are introduced from other regions or countries for various reasons such as aquaculture, sport fishing, or aquarium trade.
- These fishes may not be well adapted to the local environmental conditions, and their introduction can sometimes have negative impacts on the native species and ecosystems.
- Examples of exotic freshwater fishes include the popular aquarium fish species like guppies, goldfish, and tropical cichlids.
- These fishes are often bred in captivity and can be found in home aquariums worldwide.
In conclusion, Catla and Rohu are examples of indigenous freshwater fishes native to the Indian subcontinent. They are highly valued food fishes and have been traditionally farmed and consumed in the region for a long time.

Definition:

Sericulture refers to the rearing of silk worms for the production of silk.

Explanation:

Sericulture is an ancient practice that involves the cultivation of silkworms and the production of silk fibers from their cocoons. Here is a detailed explanation of each option and why they are incorrect:

A. Apiculture:
- Apiculture is the practice of beekeeping for the production of honey and other bee products.
- It is unrelated to the rearing of silk worms for silk production.
B. Sericulture:
- Sericulture is the correct answer.
- It involves the rearing of silk worms (Bombyx mori) in controlled environments.
- The silk worms are fed mulberry leaves until they spin their cocoons.
- The cocoons are then harvested, and the silk fibers are extracted for various uses, including fabric production.
C. Pisciculture:
- Pisciculture refers to the breeding and rearing of fish in controlled environments.
- It has no connection to the production of silk.
D. Horticulture:
- Horticulture is the science and art of cultivating plants, including fruits, vegetables, flowers, and ornamental plants.
- While horticulture is related to agriculture, it does not involve silk production.
In conclusion,
sericulture is the correct term for the rearing of silk worms for the production of silk.

The study of fish culture is called Pisciculture.

Explanation:

The term "fish culture" refers to the breeding, rearing, and management of fish for commercial or recreational purposes. Pisciculture, also known as fish farming or aquaculture, is the scientific study and practice of fish culture.

Key Points:

• Pisciculture: The study of fish culture is called pisciculture. It involves the breeding, rearing, and management of fish in controlled environments.


• Ophiology: Ophiology is the study of snakes.


• Ichthyology: Ichthyology is the study of fish, including their classification, anatomy, behavior, and distribution.


• Herpetology: Herpetology is the study of reptiles and amphibians.

In conclusion:

The correct answer is D. Pisciculture. It is important to note that ophiology, ichthyology, and herpetology are all related to the study of animals, but they specifically focus on snakes, fish, and reptiles/amphibians, respectively.

Explanation:

The breeds mentioned, ILS-82, HH-260, and B-77, do not correspond to any known breeds of pigs, cows, or buffaloes. Therefore, the correct answer is option C: fowl.

Reasoning:

• ILS-82, HH-260, and B-77 are not commonly recognized breeds of pigs, cows, or buffaloes.


• Typically, breed names for pigs begin with "Sw-" (e.g., Swabian-Hall), for cows with the name of the breed (e.g., Holstein), and for buffaloes with "Murrah" or "Mehsana" (e.g., Murrah buffalo).


• Therefore, it can be concluded that the breeds ILS-82, HH-260, and B-77 are most likely referring to breeds of fowl, such as chicken or turkey, as these animals have a wider variety of breed names.

Summary:

The breeds ILS-82, HH-260, and B-77 do not correspond to known breeds of pigs, cows, or buffaloes. The most likely option is that they are breeds of fowl.

Murrah is a high-yielding breed of buffalo.
Explanation:
- Murrah is a specific breed of buffalo known for its high milk production.
- It originated in Haryana, India and is widely popular in dairy farming.
- The Murrah buffaloes are recognized for their excellent milk quality, high fat content, and high milk yield.
- They are known to produce an average of 2200-2500 liters of milk per lactation period, which is significantly higher than other buffalo breeds.
- Murrah buffaloes have a strong and sturdy build, well-developed udders, and a docile temperament.
- They are often preferred by dairy farmers due to their ability to adapt to various climatic conditions and their resistance to diseases.
- The milk from Murrah buffaloes is commonly used for making various dairy products such as ghee, butter, and cheese.
- Due to their high milk production potential, Murrah buffaloes play a crucial role in meeting the growing demand for dairy products.