# Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ =

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmev(b) visible, with λ = 500 nmev(c) x-ray, with λ = 0.50 nmev

## This Post Has 4 Comments

1. camilacarrielh15 says:

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m

2. pippalotta says:

1. The frequency of green light that has a wavelength of the $5.14 \times 10^-^7 m \ is \ 5.83 \times 10^1^4 Hz.$

2. Wavelength of infrared radiation with frequency $3.85 \times 10^1^2 Hz \ is \ 7.79 \times 10^-^5 m.$

3. Energy of a photon with frequency $7.85 \times 10^1^5 is \ 5.181 \times 10^-^1^8 J.$

4. The energy of a photon having a wavelength of $6.08 \times 10^-^7m \ is \ 3.25 \times 10^-^1^9J.$

5. Wavelength of ultraviolet radiation having $2.94 \times 10^-^8J \ is \ 6.73 \times 10^-^1^8m.$

Explanation:

The equation connecting wavelength, frequency and speed of electromagnetic radiation is

c=ϑλ

1. λ = $5.14 \times 10^-^7m$

$c=3 \times 10^8 m/s$

ϑ = c/λ = $\frac{(3 \times ^8)}{(5.14 \times 10^-^7)} = 5.83 \times 10^1^4 Hz$

2. ϑ = $3.85 \times 10^1^2 Hz$

λ= c/ϑ $= \frac{ (3 \times 10^8)}{(3.85 \times 10^1^2 )} = 7.79 \times 10^-^5 m$

3. E = h ϑ

$= 6.6 \times 10^-^3^4 \times 7.85 \times 10^1^5=5.181 \times 10^-^1^8J$

4. E= hc/λ

$= \frac{ (6.6 \times 10^-^3^4 \times 3 \times 10^8)}{ (6.08 \times 10^-^7)}$

$=3.25 \times 10^-^1^9J$

5. E=$2.94 \times 10^-^8J$

λ= hc/E

$= \frac{ (6.6 \times 10^-^3^4) \times 3 \times 10^8}{(2.94 \times 10^-^8)}$

$=6.73 \times 10^-^1^8m$

3. valejuan says:

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

4. lilpetals says:

The energy $E$ of a photon is given by the following formula:

$E=h.f$(1)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the frequency has an inverse relation with the wavelength $\lambda$:

$f=\frac{c}{\lambda}$(2)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (2) in (1):

$E=\frac{hc}{\lambda}$(3)

Knowing this, let's begin with the answers:

(a) Microwave: 50.00 cm

For $\lambda=50cm=0.5m$

$E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5m}$

$E=\frac{1.24(10)^{-6}eV.m }{0.5m}$

$E=2.48(10)^{-6}eV$

(b) Visible: 500 nm

For $\lambda=500nm=500(10)^{-9}m$

$E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{500(10)^{-9}m}$

$E=\frac{1.24(10)^{-6}eV.m }{500(10)^{-9}m}$

$E=2.48 eV$

(c) X-ray: 0.5 nm

For $\lambda=0.5nm=0.5(10)^{-9}m$

$E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5(10)^{-9}m}$

$E=\frac{1.24(10)^{-6}eV.m }{0.5(10)^{-9}m}$

$E=2480 eV$

As we can see, as the wavelength decreases, the energy increases.