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A 600-nm photon strikes a metal surface and loses 30 percent of its energy to the ejected electrons. Calculate the new momentum of the photon.
1. 7.73 × 10-24 kg m s-1
2. 2.33 × 10-24 kg m s-1
3. 1.34 × 10-24 kg m s-1
4. 0.76 × 10-24 kg m s-1

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Correct Answer - Option 1 : 7.73 × 10-24 kg m s-1

The correct answer is option 1) i.e. 7.73 × 10-24 kg m s-1

CONCEPT:

  • The energy of a photon:
    • Photons have wave-like and particle-like properties. Light is a form of energy and this energy comes from the energy of the photons that constitute light. 
    • Photon is a bundle of electromagnetic energy.

The energy of a photon (E) is given by:

 \(E =\frac{hc}{λ}\)

  • Momentum of a photon: The momentum (p) of a photon is related to its wavelength (λ) as follows

p = \(\frac{h}{λ}\)

Where h is Planck's constant (6.63 × 10-34 J s)

CALCULATION:

Given that:

Wavelength of the photon, λ = 600 nm = 600 × 10-9 m

We know, \(E =\frac{hc}{λ}\)

Substituting p = \(\frac{h}{λ}\) in E we get, E = pc

Since it loses 30% of its energy to the electron, the energy it will have now = 70%

If p is the initial momentum, the new momentum p' = 70% of p = 0.7p (∵ speed of light (c) is a constant)

p' = 0.7p = 0.7 × \(\frac{h}{λ}\) = 0.7 × \(\frac{6.63 × 10^{-34}}{600 × 10^{-9}}\) = 7.73 × 10-28 kg m s-1

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