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Dmitriy789 [7]
3 years ago
14

The energy of a wave will remain constant if which of the following changes are made to it?A The wavelength is cut in half, and

the speed is doubled.B The wavelength is doubled, and the amplitude is tripled.c wavelength is cut in half, and the speed is tripled. D The wavelength is doubled, and the speed is also doubled.
Chemistry
1 answer:
aliya0001 [1]3 years ago
6 0

The energy of a wave will remain constant if the wavelength is doubled and the speed is also doubled.

Option D

<u>Explanation:</u>

Based on the dual nature of waves, Planck's equation states that the energy of the wave is directly proportional to the frequency of the wave. The Planck constant is termed as the proportionality constant.

So, E = hv

It is known that frequency is the ratio of speed of light to wavelength of wave, so the energy equation can be written as

E = \frac{hc}{lambda}

Thus, energy is inversely proportional to the wavelength of wave and directly proportional to the speed of wave. So in order to keep the energy constant, both the wavelength and the speed should be doubled as shown below.

Let  c = 2c and λ = 2λ, then the new energy will be

E'=\frac{hc}{lambda}

Since, c = 2c and lambda = 2 lambda, E' = 2hc/2lambda = E

So the wavelength is doubled and the speed is also doubled to keep the energy of the wave constant.

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Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
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Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

75 \times \frac{1}{2} =48.2

34.5 \times \frac{1}{2} =72.3

17.25 \times \frac{1}{2} =96.4

8.625\times \frac{1}{2} =120.5

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

A=A_{0} \cdot \frac{1}{2^{n}}

Where

A_0  is the initial sample amount  

n = the number of half-lives that pass in a given period of time.

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a. mol O₂=0.5

b. volume O₂ = 25 cm³

c. i. the total volume of the two reactants = 75 cm³

c. ii. the volume of nitrogen dioxide formed = 50 cm³

<h3>Further explanation</h3>

Reaction

2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)

a.

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From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :

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b.

From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles  

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\tt \dfrac{1}{2}\times 50~cm^3=25~cm^3

c.

i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³

ii. the volume of nitrogen dioxide formed :

mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³

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