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3 years ago

A. Describe the photoelectric effect and explain why it could not be explained by Newtonian physics.

2 answers:

8 0

a) This effect results in emission of electrons out of the lighted area of metals. According to Newtonian mechanics, energy is continuous (may take any value), but photoelectric effect implies using discrete energy ideas.

b) 'Quantum' means 'part' and is being used in physics fir describing light. Light consists of separate particles. This theory fits well what we observe in photoelectric effect: one light particle smashes one electron and it leaves surface of a metal.

c) Quantum mechanics describes everything as wave and particle at a time and it is called dualism. Electrons have separate quantified energetic levels and they go up/down consuming/emitting light waves. The lines in the spectrum fit these energetic levels.

Leno4ka [110]3 years ago

3 0

Explanation:

(a)

The photoelectric effect is the phenomenon in which the light of the particular frequency incidents on the material. Then the emission of the electrons from the surface of the material occurs.

This phenomenon could not be explained by Newtonian physics.

In Newtonian physics, the energy is not discrete. In quantum mechanics, the energy is discrete. This is the main why the photoelectric effect could not be explained by Newtonian physics.

(b)

Light consists of photons. The photon is a packet of energy. It is also called quanta. The energies of the photons are quantized.

When a photon strikes on the surface of metal then the energy of photon is absorbed by an electron in the metal so that it may eject from the surface. This phenomenon is called the photoelectric effect.

(c)

In quantum mechanics, wave-particle duality concept is used to explain the wave-particle nature of the light. Light behaves as particle as well as wave. It shows both nature. The photoelectric phenomenon shows the particle nature of the light. It acts as a particle when it hits the surface of the metal.

In line spectra, the electron is excited to an energy level. In this case energy is transferred from photon to electron. There is a collision between photon and electron. The change in momentum will occur. It shows the particle nature of the light.

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Answer:

When the sound wave enters the water its wavelength increase.

Explanation:

Given:

Speed of sound in air $v = 343 \frac{m}{s}$

Speed of sound in fresh water $v' = 1482 \frac{m}{s}$

We know that when wave travel into one medium to another medium frequency doesn't change.

Velocity is given by,

$v = f \lambda$

In above equation frequency is constant,

So when sound wave enters the water wavelength increase because speed sound is increase in water as compare to air.

Therefore, when the sound wave enters the water its wavelength increase.

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3 years ago

A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then

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Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

$v^{2}-u^{2}= 2as$

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$v^{2}-(0)^{2}= 2(1.44)(30)$

$v= 9.3 m^{-1}$

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

$s=ut+\frac{(1)}{2}at^2$

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$30=(0)t+\frac{(1)}{2}(1.44)t^2$

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

$T=6.45+\frac{s'}{v}$

Put s= 30 m, $v= 9.3 m^{-1}$ and s'= 100 m.

$T=6.45+\frac{(100)}{9.3}$

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

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3 years ago

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

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The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

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The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

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