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LenKa [72]
3 years ago
10

How did Rutherford's experiments demonstrate that Thomson's model of the atom was incorrect?

Physics
2 answers:
chubhunter [2.5K]3 years ago
8 0
Rutherford's Gold foil experiment involved the firing of atoms at a piece of gold foil only a few atoms thick. Rutherford expected that the particles would go straight through, but some were deflected. This experiment proved that atoms have a dense positively charged nucleus.

Answer: d. They showed that atoms have a dense nucleus.




mestny [16]3 years ago
8 0

Answer:

d) They showed that atoms have a dense nucleus.

Explanation:

Thomson's  model  predicted that the atom is composed of negatively charged electrons, which is surrounded by positively charged cloud. This was called Plum pudding model of the atom.

But Rutherford proved it to be incorrect . His gold foil experiment showed that the atom's central part is heavy and positively charged. The deflections of the alpha particles showed that the heavy central region is surrounded by negatively charged particles called electrons.

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Answer:

the product of mass and velocity

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3 years ago
‏Explain how an electron emitted by the photoelectric effect can have kinetic energy less than threshold energy ?
xenn [34]

Answer:

the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

          K = E - Ф

Explanation:

The photoelectric effect is the emission of electrons from the surface of a metal.

This was correctly explained by Einstein, in his explanation the energy of the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

          E = hf

          E = K + Ф

          K = E - Ф

The energy of the photons is given by the Planck relation E = hf and according to Einstein the number of joints must be added

            E = n hf

Therefore, depending on the value of this energy, the emitted electrons can have energy from zero onwards.

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3 years ago
Which one of the following weight management plans is the most effective
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A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr
Ksju [112]

Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

hf-32=57.29

hf=57.29+32

hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

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