<h2>
Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>
Explanation:
The half-life 
of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.
In this case, we are given the half life of two elements:
beryllium-13: 
beryllium-15: 
As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?
We can find it out by the following expression:
Where 
is the amount we want to find:
Finally:
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.
<span>B. If the temperature of a solid is increased, the solid expands and its volume increases.</span>
The mechanical efficiency = actual work / ideal work
So ζ = 1540 / 1600 * 100% = 96.25%
To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.
Mathematically its definition can be given as:
Where
d = Width of the slit
Angle between the beam and the source
m = Order (any integer) which represent the number of repetition of the spectrum, at this case 1 (maximum respect the wavelength)
Since the point of the theta angle for which the diffraction becomes maximum will be when it is worth one then we have to:
Applying the given relation of frequency, speed and wavelength then we will have that the frequency would be:
Here the velocity is equal to the speed of light and the wavelength to the value previously found.
Therefore the smallest microwave frequency for which only the central maximum occurs is 1.5Ghz
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