Question

a. What is the frequency of radiation that has a wavelength of 11 μm , about the size of a bacterium? b. What is the wavelength of radiation that has a frequency of 5.30×10^14 s^−1 ?c. Would the radiations in part (a) or part (b) be visible to the human eye?​d. What distance does electromagnetic radiation travel in 45.0 μs ?

Answer 1

Answer:

a)  ν = 2.7 x 10¹³  s⁻¹

b)  λ=  5.7 x 10⁻⁷ m

c)  Neither will be visible

d)  d= 13,500 m

Explanation:

The relationship ν = c/λ  where

ν = frequency of the radiation

c = speed of light = 3 x 10^8 m/s

λ = wavelength of the radiation

will be used to solve parts a), b) and c).

For part d. we know that all electromagnetic radiation travel at the speed of light so d= speed x time

a) ν = 3 x 10^8 m/s / 11 x 10 ⁻⁶ m       ( 1 μm = 10⁻6 m )

      = 2.7 x 10¹³  s⁻¹

b)  λ= c/ν = 3 x 10^8 m/s / 5.30 x 10¹⁴ s⁻¹ = 5.7 x 10⁻⁷ m

c)  The visible spectrum range is 380 to 470 nm

in a. converting to nm:

11 x 10⁻⁶ m x  10⁹ nm/m = 11,000 nm

in b. converting to nm:

5.7 x 10⁻⁷ m x  10⁹ nm/ m = 570 nm

Neither of these radiations will be visible to the human eye.

d) d= 3 x 10^8 m/s x 45 x 10⁻⁶ s = 13,500 m

Answer 2

The frequency of the radiation is 2.73 x 10¹³ Hz.

The wavelength of the radiation is 566 nm

The distance traveled by the electromagnetic radiation is 13,500 m.

The given parameters;

• wavelength of the radiation, λ = 11 μm = 11 x 10⁻⁶ m.

The frequency of the radiation is calculated as follows;

c = fλ

$f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^{8}}{11 \times 10^{-6}} \\\\f = 2.73 \times 10^{13} \ Hz$

The wavelength of the radiation is calculated as follows;

$\lambda = \frac{c}{f} \\\\\lambda = \frac{3\times 10^{8}}{5.3\times 10^{14}} \\\\\lambda = 5.66 \times 10^{-7} \ m\\\\\lambda = 566 \ \times 10^{-9} \ m\\\\\lambda = 566 \ nm$

The distance traveled by the electromagnetic radiation is calculated as;

d = vt

d = 3 x 10⁸ x 45 x 10⁻⁶

d = 13,500 m

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