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ddd [48]
3 years ago
12

CAN SOMEONE PLEASE CHECK MY ANSER PLASE

Chemistry
2 answers:
Naya [18.7K]3 years ago
8 0
<span>F they attract
F
F (Schroegener is the cloud)
T




</span>
<span>the uncertainty principle states that the position and velocity cannot both be measured,exactly, at the same time (actually pairs of position, energy and time)
uncertainty principle derives from the measurement problem, the intimate connection between the wave and particle nature of quantum objects
the change in a velocity of a particle becomes more ill defined as the wave function is confined to a smaller region
the wave nature to particles means a particle is a wave packet, the composite of many waves
many waves = many momentums, observation makes one momentum out of many
exact knowledge of complementarity pairs (position, energy, time) is impossible
complementarity also means that different experiments yield different results (e.g. the two slit experiment)
therefore, a single reality can not be applied at the quantum level
the mathematical form of the uncertainty principle relates complementary to Planck's constant
knowledge is not unlimited, built-in indeterminacy exists, but only in the microscopic world, all collapses to determinism in the macroscopic world



True



Pictures:

pic.1) true

pic 2) I got no idea









</span>
Leokris [45]3 years ago
6 0
1.True
2.False
3.B
4.A
5.True
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Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
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Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

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= De-Broglie's wavelength = ?

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Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

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