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11.Dual Nature of Radiation and matter
hard
If the total energy transferred to a surface in time $t$ is $6.48 \times 10^5 \mathrm{~J}$, then the magnitude of the total momentum delivered to this surface for complete absorption will be :
A
$2.46 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
B
$2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
C
$1.58 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
D
$4.32 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
(JEE MAIN-2024)
Solution
$\mathrm{p}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}$
Standard 12
Physics
