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A photon of wavelength 6630Å is incident on a totally reflecting surface. The momentum delivered by the photon is equal to
  • a)
    6.63 x 10—27 kg m/sec
  • b)
    2 x 10—27 kg m/sec
  • c)
    10—27 kg m/sec
  • d)
    3 x 10—27 kg m/sec
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A photon of wavelength 6630Å is incident on a totally reflecting...
Angstroms (Å) has an energy of approximately 2.97 electronvolts (eV).

This can be calculated using the formula:

E = hc/λ

Where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 joule seconds), c is the speed of light (2.998 x 10^8 meters per second), and λ is the wavelength of the photon in meters.

Converting the wavelength from Angstroms to meters:

6630 Å = 6.63 x 10^-7 meters

Plugging in the values:

E = (6.626 x 10^-34 joule seconds) x (2.998 x 10^8 meters per second) / (6.63 x 10^-7 meters)

E ≈ 2.97 electronvolts (eV)
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Community Answer
A photon of wavelength 6630Å is incident on a totally reflecting...
Λ=h/mv
mv=h/λ
=6626×10^-37/6630×10^-10
=10^-27Kg m/sec
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