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3 years ago

The “Compton effect” showed

2 answers:

8 0

The answer is the last choice, the particle nature of light. The compton effect showed that a charged electron scatters the photon results in decreased energy but increase in wavelength.

Yanka [14]3 years ago

3 0

Answer:

the particle nature of light

Explanation:

As we know that by the formula of Compton Effect we have

$\lambda' - \lambda = \frac{h}{mc}(1 - cos\theta)$

here in this formula we will say that when light of some given wavelength $\lambda$ incident on a particle then after deviating from the particle the light will change its direction by some angle $\theta$ .

And the wavelength of the particle will also changed to some different value as $\lambda'$

so here as per particle theory we will use the theory of momentum conservation as per which light of given momentum when hit a particle then it will transfer it momentum to the particle and deviated by some lesser momentum.

Due to decrease in momentum the change in wavelength of light is given by above formula

so this verify the particle nature of light

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A gamma ray and a microwave traveling in a vacuum have the same?

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They have the same speed

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3 years ago

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A hydrogen atom that has an electron in the n = 2 state absorbs a photon. What wavelength must the photon possess to send the el

Deffense [45]

Answer:

486nm

Explanation:

in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that

$\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\$

Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state

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3 years ago

A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the

stepan [7]

Answer:

m = 1.99 kg = 2 kg

Explanation:

The moment of inertia of a bicycle rim about it's center is given by the following formula:

$I = mr^{2}\\$

where,

I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²

r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2

r = 0.65 m/2 = 0.325 m

m = Mass of the Bicycle Rim = ?

Therefore,

$0.21\ kg.m^{2} = m(0.325\ m)^{2}\\m = \frac{0.21\ kg.m^{2}}{(0.325\ m)^{2}}\\$

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3 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

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3 years ago

Which of the following is a chemical change?

klemol [59]

a). Water is still H₂O after it freezes.

b). Ice is still H₂O after it melts.

c). Wire is still Cu when it's bent.

d). Paper combines with the O₂ in the air, and turns into
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3 years ago