Atoms And Nuclei

The hydrogen spectrum is characterized by different spectral series, each corresponding to electron transitions between energy levels. The shortest wavelength in each series occurs when the transition is from the higher energy levels to the lowest level of that series. These series in the order of the smallest to largest shortest wavelengths are:

1. Lyman series (ultraviolet region): Transitions are to n=1. The shortest wavelength corresponds to the transition from n=∞ to n=1.
2. Balmer series (visible region): Transitions are to n=2. The shortest wavelength corresponds to the transition from n=∞ to n=2.
3. Paschen series (infrared region): Transitions are to n=3.
4. Brackett series (infrared region): Transitions are to n=4.
5. Pfund series (far infrared region): Transitions are to n=5. The shortest wavelength corresponds to the transition from n=∞ to n=5.

Therefore, the arrangement of these series in the order of decreasing wavelength (from longest to shortest) is:
(A) Pfund series
(C) Brackett series
(B) Balmer series
(D) Lyman series
Hence, the correct order is: (A), (C), (B), (D)

The radius of a nucleus is generally given by the formula , where A is the mass number, and is a constant. Thus, the nuclear radius ratio of two elements can be expressed as follows:

For the given elements with mass numbers and , the ratio is computed as:

Therefore, the ratio of the nuclear radii is: 2 : 1

• The kinetic energy is equal to .
• The potential energy is twice the kinetic energy but negative, .

To determine the charge to mass ratio of the given particles, we analyze each particle individually:

• Proton (A): Has a charge of +e (where e is the elementary charge) and a mass of approximately 1 atomic mass unit (amu).
• α-particle (B): Consists of 2 protons and 2 neutrons, thus has a total charge of +2e and a mass of approximately 4 amu.
• Electron (C): Has a charge of -e and a mass of approximately 1/1836 amu (significant due to small mass resulting in a high charge to mass ratio).

The charge to mass ratio is given as:

• Proton:
• α-particle:
• Electron: (high magnitude, ignore sign for comparison)

Therefore, the order of charge to mass ratio by magnitude is: (C) > (A) > (B)

To determine the density ratio of two nuclei with mass numbers and , we need to use the principle that nuclear density is a property that does not significantly change from one nucleus to another. The nuclear density is primarily dependent on the nuclear forces and is consistent across different nuclei. The density ( ) of a nucleus can be expressed as:

Given that the volume of a nucleus is proportional to its mass number ( ) raised to the power of (because volume scales with the cube of the radius, and the radius is proportional to ), both the volume and mass for any nucleus scale in the same way. The nuclear density is thus:

As this shows that the density is constant for nuclei with different mass numbers, the density ratio for any two nuclei, regardless of their mass numbers and , remains 1:1.

Therefore, the answer is: 1 : 1

According to Bohr's Model, let's examine each of the given options:

1. (A) The radius of the orbiting electron is directly proportional to ‘n’:
   In Bohr's model, the radius of the nth orbit is given by the formula: . This shows that the radius is proportional to , not itself. Hence, (A) is incorrect.
2. (B) The speed of the orbiting electron is directly proportional to 1/n:
   The speed of the electron in the nth orbit can be approximated using the formula: , where is the speed in the first orbit. Therefore, the speed is indeed inversely proportional to . Thus, (B) is correct.
3. (C) The magnitude of the total energy of the orbiting electron is directly proportional to 1/n²:
   The total energy of the nth orbit is given by: , implying that energy magnitude increases with . Hence, (C) is correct.
4. (D) The radius of the orbiting electron is directly proportional to n²:
   As noted in point (A), the formula clearly shows direct proportionality to . Therefore, (D) is correct.

So, the correct answer is (D): (B), (C), and (D) only.

The problem requires identifying the particle emissions associated with changes in atomic and mass numbers during radioactive decay. Let's examine each observation and match it with the correct particle emission:

(A) No change in atomic number or mass number: This is characteristic of gamma emission, which involves the emission of photons since photons have no mass or charge.

(B) No change in mass number but the atomic number decreases by 1: This indicates positron emission, which involves a proton converting into a neutron and emitting a positron.

(C) No change in mass number but the atomic number increases by 1: This corresponds to beta decay where a neutron turns into a proton, emitting an electron (beta particle).

(D) Atomic number decreases by 2, and the mass number decreases by 4: This occurs during alpha decay, where an alpha particle (helium nucleus) is emitted.

Based on the analysis, the correct order of particles is: Photon, positron, electron, alpha particle.