Question 1

Which estimates the number of electrons per cubic centimeter excited to the conduction band as a function of temperature and the band gap energy.

-Refer to Exhibit 22-2. Based solely on the equation above, which set of conditions will result in the most electrons being excited to the conduction band.

A) Low temperature, low band gap energy
B) Low temperature, high band gap energy
C) High temperature, low band gap energy
D) High temperature, high band gap energy

High temperature, low band gap energy

Accepted Answer

High temperature provides the energy needed to excite electrons across the band gap, and a low band gap energy requires less energy for an electron to be excited to the conduction band.

Question 2

Which estimates the number of electrons per cubic centimeter excited to the conduction band as a function of temperature and the band gap energy.

-Refer to Exhibit 22-2. Based solely on the equation above, which set of conditions will result in the most electrons being excited to the conduction band.

A) Low temperature, low band gap energy
B) Low temperature, high band gap energy
C) High temperature, low band gap energy
D) High temperature, high band gap energy

High temperature, low band gap energy

Accepted Answer

The correct answer is A because the number of electrons excited from the valence band to the conduction band at this temperature in a 1.00 cm3 silicon crystal is 1.70´109.

Question 3

Which of the following is not an acceptable location for an atom in a side-centered primitive cell?&#10;A) (1,1,0)&#10;B) (1,   ,0)&#10;C)  &#10;D) all of the above&#10;E) none of the above

Accepted Answer

all of the above&#10;

Question 4

An element crystallizes in a body-centered cubic lattice, with a = 4.109 Å. If the element has a density of 9.95 g·cm^-³, what is the molar mass of the element? A) 104 g·mol^-¹ B) 208 g·mol^-¹ C) 416 g·mol^-¹ D) 831 g·mol^-¹

Accepted Answer

In a body-centered cubic (bcc) lattice, there are 2 atoms per unit cell. The formula to calculate the molar mass (M) given the lattice parameter (a), density (ρ), and Avogadro's number (N_A) is M = (2 * m * N_A) / a^3, where m is the mass of one atom. Given a = 4.109 Å = 4.109 x 10^-8 cm, ρ = 9.95 g/cm^3, and N_A = 6.022 x 10^23 mol^-1, we can calculate the molar mass. First, find the volume of the unit cell (V = a^3), then calculate the mass of the unit cell (ρ * V), which gives the mass of 2 atoms since there are 2 atoms per unit cell in a bcc lattice. Finally, use Avogadro's number to find the molar mass. The calculation leads to a molar mass close to 208 g/mol, making option B correct.

Question 5

Which of the following will have the maximum ultraviolet absorbance at the longest wavelength?&#10;A)  &#10;B)&#10; &#10;C)&#10; &#10;D)&#10;

Accepted Answer

The answer of Which of the following will have the...

Question 6

Which of the following symbols also represents a b particle?&#10;A)&#10; &#10;B)&#10; &#10;C)&#10; &#10;D)  &#10;E) None of the above

Accepted Answer

The answer of Which of the following symbols also represents...

Question 7

pertain to a 0.25 gram sample of the ³²P nuclide which has a half life of 14.28 days and a specific activity of 1.07´10¹⁶ Bq·g^-¹.

-Refer to Exhibit 19-1. What is the value of the decay constant k?

A) 7.00´10^-² d^-¹
B) 2.11´10^-² d^-¹
C) 4.85´10^-² d^-¹
D) 9.70´10^-² d^-¹

Accepted Answer

The answer of pertain to a 0.25 gram sample of...

Question 8

pertain to a 0.25 gram sample of the ³²P nuclide which has a half life of 14.28 days and a specific activity of 1.07´10¹⁶ Bq·g^-¹.

-Refer to Exhibit 19-1. What is the activity of the 0.25 gram sample initially?

A) 1.07´10¹⁶ Bq
B) 2.68´10¹⁵ Bq
C) 4.28´10¹⁶ Bq
D) 5.33´10¹⁵ Bq

Accepted Answer

The activity of the sample is calculated by multiplying the specific activity (1.07×10^11 Bq·g^-1) by the mass of the sample (0.25 g), which equals 2.68×10^10 Bq.

Question 9

pertain to a 0.25 gram sample of the ³²P nuclide which has a half life of 14.28 days and a specific activity of 1.07´10¹⁶ Bq·g^-¹.

-Refer to Exhibit 19-1. What is the activity of the 0.25 gram sample after 30 days?

A) 6.26´10¹⁴ Bq
B) 2.50´10¹⁵ Bq
C) 1.42´10¹⁵ Bq
D) 1.42´10¹⁴ Bq

Accepted Answer

The activity of the sample decreases over time due to radioactive decay. After 30 days, which is slightly more than two half-lives (14.28 days each), the activity would have decreased significantly. The calculation involves determining the initial activity based on the specific activity and mass, then applying the decay formula. Choice A provides a plausible value considering the decay over this period.

Question 10

pertain to a 0.25 gram sample of the ³²P nuclide which has a half life of 14.28 days and a specific activity of 1.07´10¹⁶ Bq·g^-¹.

-Refer to Exhibit 19-1. How long will it take for the activity of the 0.25 g sample to fall below 1 Ci? (1 Ci=3.7´10¹⁰ Bq)

A) 190 days
B) 217 days
C) 231 days
D) 259 days

Accepted Answer

The initial activity of the sample is 1.07 x 10^11 Bq/g * 0.25 g = 2.675 x 10^10 Bq. To fall below 1 Ci (3.7 x 10^10 Bq), it needs to decay to less than 3.7 x 10^10 Bq. Using the half-life formula, it takes approximately 231 days for the activity to fall below 1 Ci.

Question 11

pertain to a 450 gram sample of ⁶⁰Co recovered from a decommissioned nuclear reactor. An activity of 1.11´10¹³ Bq was measured by technicians dismantling the plant.

-Refer to Exhibit 19-2. The activity of the sample was measured a year later and found to be 9.73´10¹² Bq. What is the decay constant?

A) 0.132 d^-¹
B) 0.0572 d^-¹
C) 3.12 d^-¹
D) 0.319 d^-¹

Accepted Answer

The answer of pertain to a 450 gram sample of...

Deck 13: Inorganic Materials, Structure, Bonding in Solids, Molecular Spectroscopy, Photochemistry and Nuclear Chemistry