
Starting with a sample of pure $^{66}Cu$, 7/8 of its decays into Zn in 15 min. The corresponding half-life is
Answer
571.5k+ views
Hint: In this question, we will use the relation between the half life and the initial amount. Here by finally adding all the half lives will give us the required result. Further, we will study the basics of wave and particle nature of light, for our better understanding.
Formula used:
$\tau = \dfrac{1}{{{2^x}}}N$
Complete step by step solution:
Let us assume the initial amount of Cu given as, N.
Now, when we see the fraction decayed in first half life is given by:
$\dfrac{1}{2}N$
Now, when we see the fraction decayed in second half lives is given by:
$\dfrac{1}{4}N$
Now, when we see the fraction decayed in third half lives is given by:
$\dfrac{1}{8}N$
So, the total amount decayed in all the three half lives are given as:
$\dfrac{1}{2}N + \dfrac{1}{4}N + \dfrac{1}{8}N = \dfrac{7}{8}N$
Since, we know 3 half lives are equal to 15 minutes.
Therefore, half life of Cu is 5minutes.
Additional Information: Wave- particle duality is a concept of quantum mechanics that according to this every particle may be described as either a particle or a wave. It expresses the inability of the classical concepts of particle and wave, to fully describe the behavior of quantum scale objects.
In general, an electron in a metal has a de-Broglie wavelength in order of ~10nm. So, we observe quantum-mechanical effects in the properties of a metal when the width of the sample is around this value. The S.I unit of this wavelength is meter (m).
De-Broglie won the Nobel Prize for physics in 1929, after the wave- like behavior of matter was experimentally demonstrated in 1927.
Note: De- Broglie wavelength gives the relation between particle and wave property. If a particle is larger than its de-Broglie wavelength, or I it is interacting with other objects on a scale significantly larger than its de Broglie wavelength, then its wave- like properties are not acceptable.
Formula used:
$\tau = \dfrac{1}{{{2^x}}}N$
Complete step by step solution:
Let us assume the initial amount of Cu given as, N.
Now, when we see the fraction decayed in first half life is given by:
$\dfrac{1}{2}N$
Now, when we see the fraction decayed in second half lives is given by:
$\dfrac{1}{4}N$
Now, when we see the fraction decayed in third half lives is given by:
$\dfrac{1}{8}N$
So, the total amount decayed in all the three half lives are given as:
$\dfrac{1}{2}N + \dfrac{1}{4}N + \dfrac{1}{8}N = \dfrac{7}{8}N$
Since, we know 3 half lives are equal to 15 minutes.
Therefore, half life of Cu is 5minutes.
Additional Information: Wave- particle duality is a concept of quantum mechanics that according to this every particle may be described as either a particle or a wave. It expresses the inability of the classical concepts of particle and wave, to fully describe the behavior of quantum scale objects.
In general, an electron in a metal has a de-Broglie wavelength in order of ~10nm. So, we observe quantum-mechanical effects in the properties of a metal when the width of the sample is around this value. The S.I unit of this wavelength is meter (m).
De-Broglie won the Nobel Prize for physics in 1929, after the wave- like behavior of matter was experimentally demonstrated in 1927.
Note: De- Broglie wavelength gives the relation between particle and wave property. If a particle is larger than its de-Broglie wavelength, or I it is interacting with other objects on a scale significantly larger than its de Broglie wavelength, then its wave- like properties are not acceptable.
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