Question

The dissociation energy of a molecule is the energy required to break apart the molecule into its separate atoms. The dissociation energy for a particular molecule is 1.12 × 10-18 J. Suppose that this energy is provided by a single photon.

Determine the (a) wavelength and (b) frequency of the photon. (c) In what region of the electromagnetic

Answer 1

Answer:

(a) 1.78 * 10^(-7) m

(b) 1.69 * 10(15) Hz

(c) Ultraviolet light

Explanation:

(a) To determine the wavelength, we use one of the formulae of energy of a wave:

E = hc/λ

Where h = Planck's constant

c = speed of light

λ = wavelength.

Therefore, wavelength, λ, is:

λ = hc/E

λ = (6.626 * 10^(-34) * 3 * 10^8) / (1.12 * 10^(-18))

λ = 1.775 * 10^(-7) m

This is the wavelength of the photon.

(b) Using formula for speed of a wave, we can find the frequency:

c = λf

Where f = frequency.

f = c/λ

f = (3 * 10^8) / (1.775 * 19^(-7))

f = 1.69 * 10^(15) Hz

This is the frequency of the photon.

(c) By comparing the frequency and wavelength of the photon with the electromagnetic spectrum, the proton is in the region of ultraviolet light.

Answer 2

a. The wavelength of this photon is equal to $1.78 \times 10^{-5} \;meter$.

b. The frequency of this photon is equal to $1.69 \times 10^{13}\;Hertz$

c. Since this photon has a wavelength of $1.78 \times 10^{-5} \;meter$, it belongs to the ultraviolet region of the electromagnetic spectrum.

Given the following data:

• Energy of photon = $1.12 \times 10^{-18 }\;Joules$

Speed of light = $3 \times 10^8 \;m/s$

a. To determine the wavelength of this photon, we would use Einstein's equation for photon energy:

Mathematically, Einstein's equation for photon energy is given by the formula:

$E = hf = h\frac{v}{\lambda}$

Where:

• E is the maximum kinetic energy.

• h is Planck constant.

• f is photon frequency.

• $\lambda$ is the wavelength.

• v is the speed of light.

Making $\lambda$ the subject of formula, we have:

$\lambda = \frac{hv}{E} \\\\\lambda = \frac{6.626 \times 10^{-34}\times 3 \times 10^{8}}{1.12 \times 10^{-18 }}\\\\\lambda =\frac{1.99 \times 10^{-25}}{1.12 \times 10^{-18 }} \\\\\lambda = 1.78 \times 10^{-5} \;meter$

b. To determine the frequency of this photon:

$Frequency = \frac{v}{\lambda} \\\\Frequency = \frac{3 \times 10^{8}}{1.78 \times 10^{-5}} \\\\Frequency = 1.69 \times 10^{13}\;Hertz$

c. Since this photon has a wavelength of $1.78 \times 10^{-5} \;meter$, it belongs to the ultraviolet region of the electromagnetic spectrum.

Read more: https://brainly.com/question/16901506