Question 1:
What is the frequency and energy of a photon with a wavelength of 10 nm?
Answer:
Speed of light in vacuum , c = 3.00 × 108 m s-1
Wavelength, λ = 10 nm
= 10 × 10-9 m
$$ \begin{aligned} &\text { Frequency, } f=\frac{c}{\lambda}\\ &\begin{aligned} & =\frac{3.00 \times 10^8}{10 \times 10^{-9}} \\ & =3.0 \times 10^{16} \mathrm{~Hz} \end{aligned} \end{aligned} $$
$$ \begin{gathered} E \propto f \\ E=h f \\ \text { where } E=\text { photon energy } \\ h=\text { Planck’s constant }\left(6.63 \times 10^{-34} \mathrm{~J s)}\right. \\ f=\text { frequency of light waves } \end{gathered} $$
$$ \text { Energy, } \begin{aligned} E & =h f \\ & =\left(6.63 \times 10^{-34}\right)\left(3.0 \times 10^{16}\right) \\ & =1.99 \times 10^{-17} \mathrm{~J} \end{aligned} $$
What is the frequency and energy of a photon with a wavelength of 10 nm?
Answer:
Speed of light in vacuum , c = 3.00 × 108 m s-1
Wavelength, λ = 10 nm
= 10 × 10-9 m
$$ \begin{aligned} &\text { Frequency, } f=\frac{c}{\lambda}\\ &\begin{aligned} & =\frac{3.00 \times 10^8}{10 \times 10^{-9}} \\ & =3.0 \times 10^{16} \mathrm{~Hz} \end{aligned} \end{aligned} $$
$$ \begin{gathered} E \propto f \\ E=h f \\ \text { where } E=\text { photon energy } \\ h=\text { Planck’s constant }\left(6.63 \times 10^{-34} \mathrm{~J s)}\right. \\ f=\text { frequency of light waves } \end{gathered} $$
$$ \text { Energy, } \begin{aligned} E & =h f \\ & =\left(6.63 \times 10^{-34}\right)\left(3.0 \times 10^{16}\right) \\ & =1.99 \times 10^{-17} \mathrm{~J} \end{aligned} $$
Question 2:
How many photons are emitted per second by a 50 W green light lamp?
[Frequency of green light, f = 5.49 × 1014 Hz]
Answer:
Photon power, P = 50 W
Planck′s constant, h = 6.63 × 10-34 J s
Frequency of green light, f = 5.49 × 1014 Hz
$$ \begin{aligned} &P=n h f\\ &\text { Number of photons emitted per second, }\\ &\begin{aligned} n & =\frac{P}{h f} \\ & =\frac{50}{\left(6.63 \times 10^{-34}\right)\left(5.49 \times 10^{14}\right)} \\ & =1.37 \times 10^{20} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
How many photons are emitted per second by a 50 W green light lamp?
[Frequency of green light, f = 5.49 × 1014 Hz]
Answer:
Photon power, P = 50 W
Planck′s constant, h = 6.63 × 10-34 J s
Frequency of green light, f = 5.49 × 1014 Hz
$$ \begin{aligned} &P=n h f\\ &\text { Number of photons emitted per second, }\\ &\begin{aligned} n & =\frac{P}{h f} \\ & =\frac{50}{\left(6.63 \times 10^{-34}\right)\left(5.49 \times 10^{14}\right)} \\ & =1.37 \times 10^{20} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
Question 3:
Given that the mass of an electron is 9.11 × 10–31 kg:
(a) what is the de Broglie wavelength of an electron beam with 50 eV kinetic energy?
[1 eV = 1.60 × 10–19 J]
(b) name a phenomenon that shows the wave properties of electrons.
Answer:
(a)
Planck’s constant, h = 6.63 × 10-34 J s
Mass of an electron, m = 9.11 × 10–31 kg
Photon energy, E = 50 × 1.60 × 10–19 J
$$ \begin{aligned} &\text { de Broglie wavelength, }\\ &\begin{aligned} E & =\frac{1}{2} m v^2 \\ 2 E & =m v^2 \\ 2 m E & =m^2 v^2 \\ m^2 v^2 & =2 m E \\ m v & =\sqrt{2 m E} \end{aligned} \end{aligned} $$
$$ \begin{aligned} \lambda & =\frac{h}{m v} \\ & =\frac{h}{\sqrt{2 m E}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2\left(9.11 \times 10^{-31}\right)\left(50 \times 1.60 \times 10^{-19}\right)}} \\ & =1.74 \times 10^{-10} \mathrm{~m} \end{aligned} $$
Given that the mass of an electron is 9.11 × 10–31 kg:
(a) what is the de Broglie wavelength of an electron beam with 50 eV kinetic energy?
[1 eV = 1.60 × 10–19 J]
(b) name a phenomenon that shows the wave properties of electrons.
Answer:
(a)
Planck’s constant, h = 6.63 × 10-34 J s
Mass of an electron, m = 9.11 × 10–31 kg
Photon energy, E = 50 × 1.60 × 10–19 J
$$ \begin{aligned} &\text { de Broglie wavelength, }\\ &\begin{aligned} E & =\frac{1}{2} m v^2 \\ 2 E & =m v^2 \\ 2 m E & =m^2 v^2 \\ m^2 v^2 & =2 m E \\ m v & =\sqrt{2 m E} \end{aligned} \end{aligned} $$
$$ \begin{aligned} \lambda & =\frac{h}{m v} \\ & =\frac{h}{\sqrt{2 m E}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2\left(9.11 \times 10^{-31}\right)\left(50 \times 1.60 \times 10^{-19}\right)}} \\ & =1.74 \times 10^{-10} \mathrm{~m} \end{aligned} $$
(b) Electron diffraction