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Monochromatic light of wavelength $632.8 \mathrm{~nm}$ is produced by a helium-neon laser. The power emitted is $9.42 \mathrm{~mW}$.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Solution:
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Verified Answer
Given $\lambda=632.8 \mathrm{~nm}=632.8 \times 10^{-9} \mathrm{~m}$ power $\mathrm{P}=9.42 \mathrm{~mW}=9.42 \times 10^{-3} \mathrm{~W}$.
(a) Energy of each photon
$$
\begin{aligned}
&\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}} \\
&=3.14 \times 10^{-19} \mathrm{~J}=\frac{3.14 \times 10^{-19}}{1.6 \times 10^{-19}}=1.96 \mathrm{eV} .
\end{aligned}
$$
Momentum of photon $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $=\frac{6.6 \times 10^{-34}}{632.8 \times 10^{-9}}=1.046 \times 10^{-27} \mathrm{~kg} . \mathrm{m} / \mathrm{s}$.
(b) no. of photons /sec incident on the target
$$
\begin{aligned}
\mathrm{n} &=\frac{\text { total power }}{\text { energy of each photon }} \\
&=\frac{\mathrm{P}}{\mathrm{E}}=\frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}}=3 \times 10^{16} / \mathrm{sec} .
\end{aligned}
$$
(c) For an $\mathrm{H}$-atom to have same momentum, Speed should be
$$
\mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}_{\mathrm{H}}}=\frac{1.046 \times 10^{-27}}{1.66 \times 10^{-27}}=0.63 \mathrm{~m} / \mathrm{s}
$$
$(\because p=m v)$
(a) Energy of each photon
$$
\begin{aligned}
&\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}} \\
&=3.14 \times 10^{-19} \mathrm{~J}=\frac{3.14 \times 10^{-19}}{1.6 \times 10^{-19}}=1.96 \mathrm{eV} .
\end{aligned}
$$
Momentum of photon $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $=\frac{6.6 \times 10^{-34}}{632.8 \times 10^{-9}}=1.046 \times 10^{-27} \mathrm{~kg} . \mathrm{m} / \mathrm{s}$.
(b) no. of photons /sec incident on the target
$$
\begin{aligned}
\mathrm{n} &=\frac{\text { total power }}{\text { energy of each photon }} \\
&=\frac{\mathrm{P}}{\mathrm{E}}=\frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}}=3 \times 10^{16} / \mathrm{sec} .
\end{aligned}
$$
(c) For an $\mathrm{H}$-atom to have same momentum, Speed should be
$$
\mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}_{\mathrm{H}}}=\frac{1.046 \times 10^{-27}}{1.66 \times 10^{-27}}=0.63 \mathrm{~m} / \mathrm{s}
$$
$(\because p=m v)$
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