Question 6:
(a) Why is a large cavity with a small hole able to act as a black body?
(b) The temperature of a black body is 4500 K and it looks orange-yellow. Describe the colour changes in the black body as the body is heated to a temperature of 9000 K.
Answer:
(a)
The rays of light that enter the large cavity will undergo repeated reflections on the inner walls of the cavity. At each reflection, part of the rays are absorbed by the inner walls of the cavity. Reflections continue to occur until all the rays are absorbed and none of them can leave the cavity. Thus, the cavity acts like a black body.
(b)
As the temperature of the black body increases, the intensity of the radiation emitted increases rapidly. The intensity of the violet-blue rays increases more than the orange-yellow rays. Therefore, the black body is violet-blue at 9000 K.
(a) Why is a large cavity with a small hole able to act as a black body?
(b) The temperature of a black body is 4500 K and it looks orange-yellow. Describe the colour changes in the black body as the body is heated to a temperature of 9000 K.
Answer:
(a)
The rays of light that enter the large cavity will undergo repeated reflections on the inner walls of the cavity. At each reflection, part of the rays are absorbed by the inner walls of the cavity. Reflections continue to occur until all the rays are absorbed and none of them can leave the cavity. Thus, the cavity acts like a black body.
(b)
As the temperature of the black body increases, the intensity of the radiation emitted increases rapidly. The intensity of the violet-blue rays increases more than the orange-yellow rays. Therefore, the black body is violet-blue at 9000 K.
Question 7:
Photograph 1 shows a communication satellite in outer space. A quantum communication attempt was performed with a laser pulse of 60 mW and a wavelength of 800 nm.
(a) What is the momentum of one photon from the laser pulse?
(b) How much energy does one photon carry?
(c) What is the number of photons per second?
(d) What is the total momentum transferred by the laser pulse per second?
Answer:
(a)
$$ \text { Momentum , } \begin{aligned} p & =\frac{h}{\lambda} \\ & =\frac{6.63 \times 10^{-34}}{800 \times 10^{-9}} \\ & =8.29 \times 10^{-28} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
(b)
$$ \begin{aligned} &\text { The energy carried by each photon } E=\frac{h c}{\lambda}\\ &\begin{aligned} & =\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{800 \times 10^{-9}} \\ & =2.49 \times 10^{-19} \mathrm{~J} \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} &\text { Number of photons per second, } n\\ &\begin{aligned} P & =\frac{n h c}{\lambda} \\ P & =n\left(\frac{h c}{\lambda}\right) \\ 60 \times 10^{-3} & =n\left(2.49 \times 10^{-19}\right) \\ n & =\frac{60 \times 10^{-3}}{2.49 \times 10^{-19}} \\ & =2.41 \times 10^{17} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
(d)
$$ \begin{aligned} & \text { Total momentum per second } \\ & =\text { momentum of one photon } \times \text { number of photons per second } \\ & =8.29 \times 10^{-28} \times 2.41 \times 10^{17} \\ & =2.0 \times 10^{-10} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
Photograph 1 shows a communication satellite in outer space. A quantum communication attempt was performed with a laser pulse of 60 mW and a wavelength of 800 nm.
(a) What is the momentum of one photon from the laser pulse?
(b) How much energy does one photon carry?
(c) What is the number of photons per second?
(d) What is the total momentum transferred by the laser pulse per second?
Answer:
(a)
$$ \text { Momentum , } \begin{aligned} p & =\frac{h}{\lambda} \\ & =\frac{6.63 \times 10^{-34}}{800 \times 10^{-9}} \\ & =8.29 \times 10^{-28} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
(b)
$$ \begin{aligned} &\text { The energy carried by each photon } E=\frac{h c}{\lambda}\\ &\begin{aligned} & =\frac{\left(6.63 \times 10^{-34}\right)\left(3.00 \times 10^8\right)}{800 \times 10^{-9}} \\ & =2.49 \times 10^{-19} \mathrm{~J} \end{aligned} \end{aligned} $$
(c)
$$ \begin{aligned} &\text { Number of photons per second, } n\\ &\begin{aligned} P & =\frac{n h c}{\lambda} \\ P & =n\left(\frac{h c}{\lambda}\right) \\ 60 \times 10^{-3} & =n\left(2.49 \times 10^{-19}\right) \\ n & =\frac{60 \times 10^{-3}}{2.49 \times 10^{-19}} \\ & =2.41 \times 10^{17} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
(d)
$$ \begin{aligned} & \text { Total momentum per second } \\ & =\text { momentum of one photon } \times \text { number of photons per second } \\ & =8.29 \times 10^{-28} \times 2.41 \times 10^{17} \\ & =2.0 \times 10^{-10} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$