Soalan 11:
Apabila sebuah fotodiod disinari dengan cahaya merah (λ = 700 nm) dan cahaya biru (λ = 400 nm), tenaga kinetik maksimum fotoelektron yang terpancar oleh cahaya biru ialah 2 kali ganda daripada cahaya merah.
(a) Berapakah fungsi kerja bahan dalam fotodiod?
(b) Berapakah panjang gelombang ambang bagi bahan fotodiod tersebut?
(c) Berapakah panjang gelombang de Broglie fotoelektron yang terkeluar oleh cahaya UV
(λ = 131 nm) dari fotodiod tersebut?
Jawapan:
(a)
$$ \begin{aligned} &\text { Fungsi kerja, } W\\ &\begin{aligned} & h f=W+K \\ & \frac{h c}{\lambda}=W+K \end{aligned} \end{aligned} $$
$$ \begin{gathered} \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{700 \times 10^{-9}}=W+K \\ (\times 2) \Rightarrow 2 \times \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{700 \times 10^{-9}}=2 W+2 K \ldots \end{gathered} $$
$$ \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{400 \times 10^{-9}}=W+2 K \ldots \ldots .(2) $$
$$ \begin{aligned} &(1)-(2): \\ & W=2 \times \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{700 \times 10^{-9}} \\ &-\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{400 \times 10^{-9}} \\ &=7.10 \times 10^{-20} \mathrm{~J} \end{aligned} $$
(b)
$$ \begin{aligned} &\text { Panjang gelombang ambang, } \lambda_0\\ &\begin{aligned} \frac{h c}{\lambda_0} & =W \\ \lambda_0 & =\frac{h c}{W} \\ \lambda_0 & =\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{7.10 \times 10^{-20} \text { dari }(\mathrm{a})} \\ & =2.80 \times 10^{-6} \mathrm{~m} \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} &\text { Panjang gelombang de Broglie, } \lambda_{\mathrm{p}}\\ &\begin{aligned} \frac{h c}{\lambda} & =W+K_{\text {maks }} \\ K_{\text {maks }} & =\frac{h c}{\lambda}-W \\ & =\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{131 \times 10^{-9}}-7.10 \times 10^{-20} \\ & =1.48 \times 10^{-18} \mathrm{~J} \end{aligned} \end{aligned} $$
$$ \begin{aligned} K & =\frac{1}{2} m v^2 \\ 2 K & =m v^2 \\ 2 m K & =m^2 v^2 \\ m v & =\sqrt{2 m K} \end{aligned} $$
$$ \begin{aligned} \lambda_{\mathrm{p}} & =\frac{h}{m_e v} \\ & =\frac{h}{\sqrt{2 m_{\mathrm{e}} K}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2\left(9.11 \times 10^{-31}\right)\left(1.48 \times 10^{-18}\right)}} \\ & =4.04 \times 10^{-10} \mathrm{~m} \end{aligned} $$
Apabila sebuah fotodiod disinari dengan cahaya merah (λ = 700 nm) dan cahaya biru (λ = 400 nm), tenaga kinetik maksimum fotoelektron yang terpancar oleh cahaya biru ialah 2 kali ganda daripada cahaya merah.
(a) Berapakah fungsi kerja bahan dalam fotodiod?
(b) Berapakah panjang gelombang ambang bagi bahan fotodiod tersebut?
(c) Berapakah panjang gelombang de Broglie fotoelektron yang terkeluar oleh cahaya UV
(λ = 131 nm) dari fotodiod tersebut?
Jawapan:
(a)
$$ \begin{aligned} &\text { Fungsi kerja, } W\\ &\begin{aligned} & h f=W+K \\ & \frac{h c}{\lambda}=W+K \end{aligned} \end{aligned} $$
$$ \begin{gathered} \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{700 \times 10^{-9}}=W+K \\ (\times 2) \Rightarrow 2 \times \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{700 \times 10^{-9}}=2 W+2 K \ldots \end{gathered} $$
$$ \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{400 \times 10^{-9}}=W+2 K \ldots \ldots .(2) $$
$$ \begin{aligned} &(1)-(2): \\ & W=2 \times \frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{700 \times 10^{-9}} \\ &-\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{400 \times 10^{-9}} \\ &=7.10 \times 10^{-20} \mathrm{~J} \end{aligned} $$
(b)
$$ \begin{aligned} &\text { Panjang gelombang ambang, } \lambda_0\\ &\begin{aligned} \frac{h c}{\lambda_0} & =W \\ \lambda_0 & =\frac{h c}{W} \\ \lambda_0 & =\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{7.10 \times 10^{-20} \text { dari }(\mathrm{a})} \\ & =2.80 \times 10^{-6} \mathrm{~m} \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} &\text { Panjang gelombang de Broglie, } \lambda_{\mathrm{p}}\\ &\begin{aligned} \frac{h c}{\lambda} & =W+K_{\text {maks }} \\ K_{\text {maks }} & =\frac{h c}{\lambda}-W \\ & =\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{131 \times 10^{-9}}-7.10 \times 10^{-20} \\ & =1.48 \times 10^{-18} \mathrm{~J} \end{aligned} \end{aligned} $$
$$ \begin{aligned} K & =\frac{1}{2} m v^2 \\ 2 K & =m v^2 \\ 2 m K & =m^2 v^2 \\ m v & =\sqrt{2 m K} \end{aligned} $$
$$ \begin{aligned} \lambda_{\mathrm{p}} & =\frac{h}{m_e v} \\ & =\frac{h}{\sqrt{2 m_{\mathrm{e}} K}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2\left(9.11 \times 10^{-31}\right)\left(1.48 \times 10^{-18}\right)}} \\ & =4.04 \times 10^{-10} \mathrm{~m} \end{aligned} $$
