42 = f = 1.02 × 10¹⁶ s⁻¹
43 = λ= 1.1 × 10⁻¹² m.
44 = Average atomic mass of Z = 19.864 amu.
45 = Average atomic mass of Mg = 24.323 amu.
46= Electronic configuration:
P₁₅ = 1s² 2s² 2p⁶ 3s² 3p³
Explanation:
Q 42 = What is the frequency of ultraviolet light with wavelength 2.94 × 10⁻⁸ m.
Given data:
wavelength of radiation = 2.94 × 10⁻⁸ m.
Speed of light = 3 × 10⁸ m/s
Frequency = ?
Solution:
Formula:
speed or velocity = wavelength × frequency
c = λ × f
f = c/ λ
f = 3 × 10⁸ m/s /2.94 × 10⁻⁸ m.
f = 1.02 × 10¹⁶ s⁻¹
43 = what is the wavelength of gamma ray with the frequency 2.73 × 10²⁰ Hz.
Given data:
wavelength of radiation = ?
Speed of light = 3 × 10⁸ m/s
Frequency = 2.73 × 10²⁰ Hz.
Solution:
Formula:
speed or velocity = wavelength × frequency
c = λ × f
λ = c / f
λ = 3 × 10⁸ m/s /2.73 × 10²⁰ s⁻¹
λ= 1.1 × 10⁻¹² m.
So the wavelength of gamma radiation is 1.1 × 10⁻¹² m.
44= Consider an element Z that has two natural occuring elements with the following percent abundances: the isotope with the mass number of 19.0 is 56.8% abundant and the isotope with mass number of 21 is 43.2% abundant. what is average atomic mass of elements Z.
Given data:
Abundance of 1st isotope = 56.8%
Abundance of second isotope = 43.2%
Atomic mass of 1st isotope = 19 amu
Atomic mass of second isotope = 21 amu
Average atomic mass = ?
Solution:
Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of Z = (19.0 × 56.8) + (21 ×43.2) /100
Average atomic mass of Z= 1079.2 + 907.2 / 100
Average atomic mass of Z = 1986.4 / 100
Average atomic mass of Z = 19.864 amu.
45 = What is the atomic mass of Mg if the natural occuring isotopes are Mg-24, 78.70%
Mg-25, 10.13%
Mg-26, 11.17%
Given data:
Abundance of Mg²⁴ = 78.70%
Abundance of Mg²⁵ = 10.13%
Abundance of Mg²⁶ = 11.17%
Average atomic mass = ?
Solution:
Average atomic mass of Mg = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) +( abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of Mg = (24 × 78.70) + (25× 10.13) + (26 × 11.17) /100
Average atomic mass of Mg= 1888.8 + 253.25 + 290.42 / 100
Average atomic mass of Mg = 2432.47 / 100
Average atomic mass of Mg = 24.323 amu.
Q = 46= What is electronic configuration for phosphorus?
Electronic configuration:
P₁₅ = 1s² 2s² 2p⁶ 3s² 3p³
Abbreviated electronic configuration:
P₁₅ = [Ne] 3s² 3p³
Properties and uses;
Phosphorus is the member of nitrogen family.
It is multivalent nonmetal.
It is present in the group fifteen.
It have five valance electrons.
Its atomic number is fifteen and atomic mass is 31.
Its melting point is 44.1 °C
Its boiling point is 280 °C.
Phosphoric acid is used in industries such as fertilizer industry.
Phosphates are used in steel, glasses, sodium lamp and also in military applications.
It is also used in tooth paste and detergents.
