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A photon has wavelength $3 \mathrm{~nm}$, then its momentum and energy respectively will be
$$
\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=\text { velocity of light }=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right]
$$
Options:
$$
\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=\text { velocity of light }=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right]
$$
Solution:
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Verified Answer
The correct answer is:
$2.21 \times 10^{-25} \mathrm{~kg} \mathrm{~ms}^{-1} ; 6.63 \times 10^{-17} \mathrm{~J}$
Momentum, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.63 \times 10^{-34}}{3 \times 10^{-9}}=2.21 \times 10^{-25} \mathrm{~kg} \mathrm{~ms}^{-1}$
Energy, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\mathrm{pc}=2.21 \times 10^{-25} \times 3 \times 10^8=6.63 \times 10^{-17} \mathrm{~J}$
Energy, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\mathrm{pc}=2.21 \times 10^{-25} \times 3 \times 10^8=6.63 \times 10^{-17} \mathrm{~J}$
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