Express the standard form: (i) radius of the earth= 6 400 000 m (ii) radius of hydrogen atom = 0.000 000 000 0529

I) 6.4 × 10^(6) m

II) 5.29 × 10^(-11) m

Explanation:

Standard form is how we write very large number in such a way that it is easier to read.

Thus;

I) radius of the earth = 6 400 000 m

What we will now have is;

6.4 × 10^(6) m

II) radius of hydrogen atom = 0.000 000 000 0529 m

What we will have is;

5.29 × 10^(-11) m

i. 6.4 × 10⁶ m.

ii. 5.29 × 10⁻¹¹ m.

Explanation:

To express a number in standard form, we write it in the form a × 10ⁿ where 1 ≤ a < 10 and n is an integer.

(i) radius of the earth= 6 400 000 m

To express the radius of the earth = 6 400 000 m in standard form, we count the number of decimal places after the first digit. We have 6 decimal places after digit 6. So, we put a decimal point between digit 6 and 4. So, we have 6.4. Since we have 6 decimal places after the six digit, the exponent of our base which is 10 is 6.

So, writing 6 400 000 in the form in a × 10ⁿ, we have a = 6.4 and n = 6. So, in standard form we have 6.4 × 10⁶ m.

So, 6 400 000 =  6.4 × 10⁶ m in standard form.

(ii) radius of hydrogen atom = 0.000 000 000 0529 m

To express the radius of hydrogen atom = 0.000 000 000 0529 m in standard form, we count the number of decimal places after the decimal point including the first non-zero digit. We have 11 decimal places after the decimal point up to and including the first non-zero digit which is digit 5. So, we put a decimal point after the first non-zero digit,5. So, we have 5.29. Since we have 11 decimal places after the decimal point, the exponent of our base which is 10 is -11. The exponent is negative since the digit places are taken after the decimal point.

So, writing 0.000 000 000 0529 m in the form in a × 10ⁿ, we have a = 5.29 and n = -11. So, in standard form we have 5.29 × 10⁻¹¹ m.

So,  0.000 000 000 0529 m = 5.29 × 10⁻¹¹ m in standard form.

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.