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arlik [135]
3 years ago
8

Bohr’s atomic model differed from Rutherford's because it explained that electrons exist in specified energy levels surrounding

the nucleus. electrons are embedded within the rest of the atom like plum pudding. an atom is an indivisible sphere. electrons circle the nucleus.
Physics
2 answers:
zubka84 [21]3 years ago
8 0

Answer:

A

Explanation:

edg 2020

insens350 [35]3 years ago
3 0

Answer:

electrons exist in specified energy levels

Explanation:

In its gold-foil scattering with alpha particles, Rutherford proved that the plum-pudding model of the atom theorised by Thomson was wrong.

From his experiment, Rutherford inferred that the atom actually consists of a very small nucleus, where all the positive charge is concentrated, and the rest of the atom is basically empty, with the electrons (negatively charged) orbiting around the nucleus at very large distance.

However, Rutherford did not specify anything about the orbits of the electrons. Later, Bohr predicted that the electrons actually orbit the nucleus in specific orbits, each orbit corresponding to a specific energy level. Bohr's model found confirmation in the observation of the emission spectrum lines: when an electron in one of the higher energy level jumps down into an orbit with lower energy, the atom emits a photon which has an energy exactly equal to the difference in energy between the two orbits (and this energy of the photon corresponds to a precise wavelength).

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To Find :

Which element accumulates in the environment due to the use of  fertilizers.

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When soil became waterlogged, soil organisms take the oxygen they need from nitrates, leaving the nitrogen in a gaseous form which escapes into the air.

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6 0
2 years ago
A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

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Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

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Equating initial and the final momenta we get

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Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0
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Answer:

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Explanation:

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marusya05 [52]

Explanation:

hope it's help you ok have a good day

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