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statuscvo [17]
3 years ago
10

How did the gold foil experiment lead to the discovery of the nucleus?

Chemistry
1 answer:
makvit [3.9K]3 years ago
5 0

Answer:

Rutherford established the nuclear theory of the atom with his gold-foil experiment. When he shot a beam of alpha particles at a sheet of gold foil, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.

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The stability of a nucleus is determined by the ratio of electrons to protons.
Lera25 [3.4K]
B. False
Stability is determined by the ratio of neutrons and protons. Electrons are not in nucleus.
5 0
3 years ago
Inside stars in space, the nuclei of hydrogen atoms fuse together to form helium atoms. This represents which form of energy?
mixas84 [53]

Explanation:

When two small nuclei combine together to form a large nuclei then it is known as nuclear fusion.

When nuclei of two hydrogen atoms fuse together then it results in the formation of a helium atom along with the release of lot of energy. This energy is nuclear energy.

This nuclear reaction is as follows.

      ^{2}_{1}H + ^{3}_{1}H \rightarrow ^{4}_{2}He + ^{1}_{0}n + Energy

Thus, we can conclude that nuclear fusion represents nuclear energy.


6 0
3 years ago
Read 2 more answers
What happens when you put magnesium carbonate and diluted water
Rina8888 [55]

Answer:

Magnesium carbonate doesn't dissolve in water, only acid, where it will effervesce (bubble).

Explanation:

 An insoluble metal carbonate reacts with a dilute acid to form a soluble salt. Magnesium carbonate, a white solid, and dilute sulfuric acid react to produce magnesium sulfate. Colourless magnesium sulfate heptahydrate crystals are obtained from this solution.

8 0
3 years ago
A soccer player kicks a ball 6.5 meters. How much time is needed for the ball to travel this distance
Ahat [919]
It is 67 I think it is
8 0
3 years ago
The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding
slega [8]

Answer:

Binding\ energy=43.43\times 10^{-20}\ J

Explanation:

Using the expression for the photoelectric effect as:

E=h\nu_0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

\nu_0=\frac {c}{\lambda_0}

Applying the equation as:

\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

Given, \lambda=4.00\times 10^{-7}\ m

\frac {hc}{\lambda_0} is the binding energy or threshold energy

\frac {1}{2}\times m\times v^2 is the kinetic energy of the electron emitted.  = 6.26\times 10^{-20}\ J

Thus, applying values as:

\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy

\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

Binding\ energy=43.43\times 10^{-20}\ J

5 0
3 years ago
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