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MrRa [10]
2 years ago
7

How as bohr's atomic model similar to Rutherford's model

Physics
2 answers:
zysi [14]2 years ago
6 0

Answer:

It described a nucleus surrounded by a large volume of space

Explanation:

In 1911 Rutherford, based on the results of his famous gold foil experiment, proposed that the atoms are made of positively charged and highly denser nucleus, around which the negatively charged and weightless electrons are revolving.

But he failed to provide the reason why the electrons are not continuously emitting electromagnetic radiation, since according to classical physics charged particle in circular motion emits radiation.

in 1913 Bohr improved this model by the quantization of orbits.

According to his theory electrons are revolving in stable orbits at particular radius and they emit radiation only when the move from one orbit to another.

So description of a nucleus surrounded by a large volume of space is the similarity to Rutherford’s model.

Lemur [1.5K]2 years ago
5 0

It described a nucleus surrounded by a large volume of space. ... Electrons travel around the nucleus in fixed energy levels with energies that vary from level to level. C) Electrons travel around the nucleus in fixed energy levels with equal amounts of energy.

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Hey! Can someone help with this question? Thx :)
Schach [20]
<span>A. Chemical energy to chemical energy</span>
4 0
2 years ago
Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod
Dima020 [189]

Answer:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>

Explanation:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.

<u>We can prove this by the equation of heat for the two bodies:</u>

<em>According to given condition,</em>

\Delta T_1=\Delta T_2

\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}

<em>when there is no heat loss from the system of two bodies then </em>Q_1=Q_2

\frac{1}{m.c_1} =\frac{1}{m.c_2}

\Rightarrow c_1=c_2

  • Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

  • when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

  • the mass of colder object is half the mass of the hotter object while their specific heat is same.
3 0
2 years ago
Three charges are arranged as shown in the picture above. Find the magnitude and direction of the electrostatic force on the 6 n
bagirrra123 [75]

Answer:

Bot Nm

Explanation:

jdjdmxjd mdjdcj jdsdj jedidj jddj

6 0
2 years ago
The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 1 mph , toward a group of small children. Seein
iogann1982 [59]

Answer:

0.5 mph in the opposite direction

Explanation:

m_1 = Mass of Cadillac = 1000 kg

v_1 = Velocity of Cadillac = 1 mph

m_2 = Mass of Volkswagen = 2000 kg

v_2 = Velocity of Volkswagen

In order to know the speed the system must have the momentum exchange

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=0\\\Rightarrow v_2=-\dfrac{m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{1000\times 1}{2000}\\\Rightarrow v_2=-0.5\ mph

The speed of the impact is given by 0.5 mph in the opposite direction

8 0
3 years ago
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

6 0
3 years ago
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