Chapter 22, Problem 26QAP

To determine

(A)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{4.14\times {10}^{15}}=7.25\times {10}^{-8}m,\text{ultra violet light}$

To determine

(B)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{7\times {10}^{14}}=4.29\times {10}^{-7}m,\text{purple visible light}$

To determine

(C)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{8\times {10}^{16}}=3.75\times {10}^{-9}m,\text{x ray}$

To determine

(d)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{3\times {10}^{13}}=1\times {10}^{-5}m,\text{infrared light}$

To determine

(E)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{9\times {10}^{12}}=3.33\times {10}^{-5}m,\text{infrared light}$

To determine

(F)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{3.44\times {10}^{17}}=8.72\times {10}^{-10}m,\text{x ray}$

To determine

(G)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{8.23\times {10}^{15}}=3.65\times {10}^{-8}m,\text{ultra violet light}$

To determine

(H)

The wavelength of the electromagnetic waves with the given frequency and classify the electromagnetic radiation of each (x ray, radio, etc.)?

Explanation of Solution

Formula used:

The wavelength of the electromagnetic waves is calculated by, $\lambda =\frac{c}{f}$

Here, c is speed of light

Calculation:

$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{6\times {10}^{15}}=5\times {10}^{-8}m,\text{ultra violet light}$