Answer :
Sure! Let's go through each question step-by-step to understand the solutions:
Question 35:
When alpha particles pass through a thin metal foil, the observation that only one out of ten thousand rebounds led to a significant conclusion. This suggests that the mass and positive charge of the atom are concentrated in a small region. This region is the nucleus. Hence, the conclusion is:
- Positively charged particles are concentrated at the center of the atom.
- Answer: d
Question 36:
The canal ray experiment, also known as the discovery of positive rays, led to the identification of protons. Protons are positively charged particles present in the atom.
- Answer: a
Question 37:
The energy difference between two adjacent electron shells decreases with increasing shell number. The minimum difference in energy levels among the given choices is between shells M and N.
- Answer: c
Question 38:
Assertion A states that an electron in the inner orbit is more tightly bound to the nucleus. Reason B explains that the greater the absolute value of the energy of an electron, the more tightly the electron is bound to the nucleus. Both the assertion and the reason are correct, and Reason B explains Assertion A.
- Answer: b
Question 39:
To find the percentage composition of the isotopes with mass numbers 15 and 11, given their average atomic weight is 14:
- Let x% of isotope with mass number 15 be present, and thus (100-x)% of isotope with mass number 11.
- The equation based on average atomic mass is:
[tex]\[ 15 \times x + 11 \times (100 - x) = 1400 \][/tex]
- Solving this equation, we find:
[tex]\[ x = 75 \][/tex]
- So, the isotopes are 75% and 25% respectively.
- Answer: c
Question 40:
To determine the percentage abundance of isotopes with mass numbers 16 and 18, given the average atomic weight is 16.5:
- Let p% of the isotope with mass number 16 be present, hence (100-p)% of isotope with mass number 18.
- The equation for average atomic mass is:
[tex]\[ 16 \times p + 18 \times (100 - p) = 1650 \][/tex]
- Solving this equation, we determine:
[tex]\[ p = 75 \][/tex]
- Thus, the isotopic abundances are 75% and 25% respectively.
- Answer: a
I hope this detailed explanation helps you understand each solution clearly!
Question 35:
When alpha particles pass through a thin metal foil, the observation that only one out of ten thousand rebounds led to a significant conclusion. This suggests that the mass and positive charge of the atom are concentrated in a small region. This region is the nucleus. Hence, the conclusion is:
- Positively charged particles are concentrated at the center of the atom.
- Answer: d
Question 36:
The canal ray experiment, also known as the discovery of positive rays, led to the identification of protons. Protons are positively charged particles present in the atom.
- Answer: a
Question 37:
The energy difference between two adjacent electron shells decreases with increasing shell number. The minimum difference in energy levels among the given choices is between shells M and N.
- Answer: c
Question 38:
Assertion A states that an electron in the inner orbit is more tightly bound to the nucleus. Reason B explains that the greater the absolute value of the energy of an electron, the more tightly the electron is bound to the nucleus. Both the assertion and the reason are correct, and Reason B explains Assertion A.
- Answer: b
Question 39:
To find the percentage composition of the isotopes with mass numbers 15 and 11, given their average atomic weight is 14:
- Let x% of isotope with mass number 15 be present, and thus (100-x)% of isotope with mass number 11.
- The equation based on average atomic mass is:
[tex]\[ 15 \times x + 11 \times (100 - x) = 1400 \][/tex]
- Solving this equation, we find:
[tex]\[ x = 75 \][/tex]
- So, the isotopes are 75% and 25% respectively.
- Answer: c
Question 40:
To determine the percentage abundance of isotopes with mass numbers 16 and 18, given the average atomic weight is 16.5:
- Let p% of the isotope with mass number 16 be present, hence (100-p)% of isotope with mass number 18.
- The equation for average atomic mass is:
[tex]\[ 16 \times p + 18 \times (100 - p) = 1650 \][/tex]
- Solving this equation, we determine:
[tex]\[ p = 75 \][/tex]
- Thus, the isotopic abundances are 75% and 25% respectively.
- Answer: a
I hope this detailed explanation helps you understand each solution clearly!