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
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
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
1. The frequency of green light that has a wavelength of the [tex]5.14 \times 10^-^7 m \ is \ 5.83 \times 10^1^4 Hz.[/tex]
2. Wavelength of infrared radiation with frequency [tex]3.85 \times 10^1^2 Hz \ is \ 7.79 \times 10^-^5 m.[/tex]
3. Energy of a photon with frequency [tex]7.85 \times 10^1^5 is \ 5.181 \times 10^-^1^8 J.[/tex]
4. The energy of a photon having a wavelength of [tex]6.08 \times 10^-^7m \ is \ 3.25 \times 10^-^1^9J.[/tex]
5. Wavelength of ultraviolet radiation having [tex]2.94 \times 10^-^8J \ is \ 6.73 \times 10^-^1^8m.[/tex]
Explanation:
The equation connecting wavelength, frequency and speed of electromagnetic radiation is
c=ϑλ
1. λ = [tex]5.14 \times 10^-^7m[/tex]
[tex]c=3 \times 10^8 m/s[/tex]
ϑ = c/λ = [tex]\frac{(3 \times ^8)}{(5.14 \times 10^-^7)} = 5.83 \times 10^1^4 Hz[/tex]
2. ϑ = [tex]3.85 \times 10^1^2 Hz[/tex]
λ= c/ϑ [tex]= \frac{ (3 \times 10^8)}{(3.85 \times 10^1^2 )} = 7.79 \times 10^-^5 m[/tex]
3. E = h ϑ
[tex]= 6.6 \times 10^-^3^4 \times 7.85 \times 10^1^5=5.181 \times 10^-^1^8J[/tex]
4. E= hc/λ
[tex]= \frac{ (6.6 \times 10^-^3^4 \times 3 \times 10^8)}{ (6.08 \times 10^-^7)}[/tex]
[tex]=3.25 \times 10^-^1^9J[/tex]
5. E=[tex]2.94 \times 10^-^8J[/tex]
λ= hc/E
[tex]= \frac{ (6.6 \times 10^-^3^4) \times 3 \times 10^8}{(2.94 \times 10^-^8)}[/tex]
[tex]=6.73 \times 10^-^1^8m[/tex]
(a) [tex]2.5\cdot 10^{-6}eV[/tex]
The energy of a photon is given by:
[tex]E=\frac{hc}{\lambda}[/tex]
where
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
[tex]c=3\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda[/tex] is the wavelength
For the microwave photon,
[tex]\lambda=50.00 cm = 0.50 m[/tex]
So the energy is
[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J[/tex]
And converting into electronvolts,
[tex]E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV[/tex]
(b) [tex]2.5 eV[/tex]
For the energy of the photon, we can use the same formula:
[tex]E=\frac{hc}{\lambda}[/tex]
For the visible light photon,
[tex]\lambda=500 nm = 5 \cdot 10^{-7}m[/tex]
So the energy is
[tex]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[/tex]
And converting into electronvolts,
[tex]E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV[/tex]
(c) [tex]2500 eV[/tex]
For the energy of the photon, we can use the same formula:
[tex]E=\frac{hc}{\lambda}[/tex]
For the x-ray photon,
[tex]\lambda=0.5 nm = 5 \cdot 10^{-10}m[/tex]
So the energy is
[tex]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[/tex]
And converting into electronvolts,
[tex]E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV[/tex]
Answers:
The energy [tex]E[/tex] of a photon is given by the following formula:
[tex]E=h.f[/tex](1)
Where:
[tex]h=4.136(10)^{-15}eV.s[/tex] is the Planck constant
[tex]f[/tex] is the frequency
Now, the frequency has an inverse relation with the wavelength [tex]\lambda[/tex]:
[tex]f=\frac{c}{\lambda}[/tex](2)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum
Substituting (2) in (1):
[tex]E=\frac{hc}{\lambda}[/tex](3)
Knowing this, let's begin with the answers:
(a) Microwave: 50.00 cm
For [tex]\lambda=50cm=0.5m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{0.5m}[/tex]
[tex]E=2.48(10)^{-6}eV[/tex]
(b) Visible: 500 nm
For [tex]\lambda=500nm=500(10)^{-9}m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{500(10)^{-9}m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{500(10)^{-9}m}[/tex]
[tex]E=2.48 eV[/tex]
(c) X-ray: 0.5 nm
For [tex]\lambda=0.5nm=0.5(10)^{-9}m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5(10)^{-9}m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{0.5(10)^{-9}m}[/tex]
[tex]E=2480 eV[/tex]
As we can see, as the wavelength decreases, the energy increases.