Metallic gold crystallizes in a face-centered cubic lattice, with one au atom per lattice point. if the edge length of the unit
1 answer:
The metallic radius (r) of au is 144.2pm
A face-centered cubic unit cell structure is made up of atoms arranged in a cube with six additional whole atoms placed in the centre of each cube face and a fraction of an atom at each of the cube's four corners. Eight additional unit cells share the atoms at the cube's corner.
Atoms in an FCC crystal are in contact along the unit cell's diagonal. 4r = √2 a, where r is the radius of an atom and a is the edge length of a unit cell.
Putting the values in the above expression,we get
r = √2×408 / 4
r =
r =
r = 144.2 pm
Therefore, The metallic radius (r) of au is 144.2pm
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Answer:
the required distance is 89.125 m
Explanation:
Given the data in the question;
we know that, sound intensity B in decibels of sound is;
β(dB) = 10log₁₀( /
)
where intensity = power / area carried by wave
= 10⁻¹² W/m² { minimum threshold intensity }
Now,
intensity = power / area carried by wave = P/A = P/4πr² { spherical }
given that; β = 80.0 dB and P = 10 W
so
β(dB) = 10log₁₀( /
)
we substitute
80 = 10log₁₀( P / 4πr²× )
80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )
8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )
8 = 1 - log₁₀( 4πr²× 10⁻¹² )
8 - 1 = -log₁₀( 4πr²× 10⁻¹² )
7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )
7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]
7 = -[ -10.9 + log₁₀( r² ) ]
7 = 10.9 - log₁₀( r² )
-log₁₀( r² ) = 7 - 10.9
-log₁₀( r² ) = - 3.9
log₁₀( r² ) = 3.9
2log₁₀r = 3.9
log₁₀r = 3.9 /2
log₁₀r = 1.95
r = 89.125 m
Therefore, the required distance is 89.125 m
The wavelength of a wave is obtained by taking the ratio of wave speed and frequency.
The wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.
What is frequency of a wave?
The number of oscillations completed by a wave in one second is known as the frequency of a wave. It is expressed as the ratio of the velocity of the wave to its wavelength.
Given data-
The wavelength of the wave is, .
The velocity of the wave is, v = 9.8 m/s.
The mathematical expression for the frequency of the wave is,
Solving as,
Now, with constant frequency and the double magnitude of velocity (v' = 2 × 9.8 = 19.6 m/s). The wavelength of the heated wave is calculated as,
here,
is the wavelength of the heated wave.
Solving as,
Thus, we can conclude that the wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.
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