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A photon of wavelength 5030 \mathrm{~A}^{\circ} is incident on a totally reflecting surface. The momentum delivered by the photon is equal to

Option: 1

6.63 \times 10^{-27} \mathrm{kgm} / \mathrm{s}


Option: 2

2 \times 10^{-27} \mathrm{kgm} / \mathrm{s}


Option: 3

10^{-27} \mathrm{kgm} / \mathrm{s}


Option: 4

None \, \, of \, \, these


Answers (1)

best_answer

The momentum of the incident radiation is given as  \mathrm{P}=\frac{\mathrm{h}}{\lambda}

When the light is totally reflected normal to the surface the direction
of the ray is reversed. That means it reverses the direction of its
momentum without changing its magnitude.

\Rightarrow \quad Change in momentum has a magnitude 

\Delta \mathrm{P}=2 \mathrm{P}=\frac{2 \mathrm{~h}}{\lambda}

\Rightarrow \quad \Delta \mathrm{P}=\frac{2\left(6.63 \times 10^{-34} \mathrm{~J}-\mathrm{sec}\right)}{\left(6630 \times 10^{-10} \mathrm{~m}\right)}

\Rightarrow \quad \Delta \mathrm{P}=2 \times 10^{-27} \mathrm{kgm} / \mathrm{s}

 

Posted by

Pankaj Sanodiya

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