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(D) The momentum $p$ of a photon is related to its wavelength $\lambda$ by the de Broglie relation: $p = \frac{h}{\lambda}$.Given the wavelength $\lam

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$10^{12} \, h$

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(D) The momentum $p$ of a photon is related to its wavelength $\lambda$ by the de Broglie relation: $p = \frac{h}{\lambda}$.Given the wavelength $\lambda = 0.01 \, \mathring{A} = 0.01 	imes 10^{-10} \, m = 10^{-12} \, m$.Substituting the value of $\lambda$ into the momentum formula:$p = \frac{h}{10^{-12} \, m} = 10^{12} \, h \, kg \cdot m/s$.Therefore,the momentum of the photon is $10^{12} \, h$.

If $h$ is Planck's constant in $SI$ system,the momentum of a photon of wavelength $0.01 \, \mathring{A}$ is:

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