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3 years ago

Why must indirect evidence be used to study the structure of atoms?

1 answer:

8 0

Indirect evidence must be used to study the structure of atoms, because the direct evidence affects the position and energy of the particles of the atomic structure.

For example, imagine you use a light ray to find the location of an electron in an atomic structure, the current location of the electron you are now seeing is as a result of the light ray, and the electron has changed it's initial position to a new position as a result of the effect of the light ray.

So indirect means would probably give more accurate answers about the atomic structure than that of the direct means.

An example of the indirect means would be for example setting up a mathematical equation or model of the atomic structure and trying to provide solution for the model or equation.

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professor190 [17]

Answer:

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3 years ago

It is advised not to hold the thermometer by the bulb while reading it

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3 years ago

Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air

vova2212 [387]

Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

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Mass = (0.00118g/cm³)(0.0593 cm³)
Mass = 7 grams of HCN

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3 years ago

What can i do to recycle?

mafiozo [28]

Answer:

You could throw away the following items:

Mixed paper: Newspaper, magazine, junk mail, cardboard, frozen food packaging, wrapping paper, paper bags, etc.

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Hope this helped

8 0

2 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

3 years ago