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An electromagnetic wave of wavelengthlamdha is incident on a photosensitive surface of negligible work function .if m mass is of photoelectron emmitted from the surface has de broglie wavelength lamfha d then?
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An electromagnetic wave of wavelengthlamdha is incident on a photosens...
Introduction:
When an electromagnetic wave of wavelength lambda is incident on a photosensitive surface, electrons may be emitted from the surface. These emitted electrons are known as photoelectrons.

Photoelectric Effect:
The photoelectric effect is the phenomenon of the emission of electrons from a photosensitive surface when it is exposed to electromagnetic radiation.

De Broglie Wavelength:
De Broglie wavelength is the wavelength associated with a moving particle. It is given by the equation lambda d = h/p, where lambda d is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle.

Explanation:
When an electromagnetic wave of wavelength lambda is incident on a photosensitive surface, electrons may be emitted from the surface. The energy of the incident electromagnetic wave is transferred to the electrons, which gain enough energy to overcome the work function of the surface and are emitted as photoelectrons.

If m mass is of photoelectron emmitted from the surface has de broglie wavelength lambda d, then we can use the equation lambda d = h/p to calculate the momentum of the photoelectron. The momentum of the photoelectron is given by p = h/lambda d.

Since the energy of the incident electromagnetic wave is transferred to the photoelectron, we can also equate the energy of the incident electromagnetic wave to the kinetic energy of the photoelectron. The kinetic energy of the photoelectron is given by KE = 1/2mv^2, where m is the mass of the photoelectron and v is its velocity.

Using the relationship between momentum and kinetic energy, p = mv, we can substitute for p in the equation p = h/lambda d to get mv = h/lambda d. Solving for v, we get v = h/(m lambda d).

Substituting for v in the equation for kinetic energy, we get KE = 1/2m(h/(m lambda d))^2 = h^2/(2m lambda d)^2.

Conclusion:
In conclusion, when an electromagnetic wave of wavelength lambda is incident on a photosensitive surface, electrons may be emitted from the surface as photoelectrons. The momentum and kinetic energy of the photoelectron can be calculated using the de Broglie wavelength and the energy of the incident electromagnetic wave.
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An electromagnetic wave of wavelengthlamdha is incident on a photosensitive surface of negligible work function .if m mass is of photoelectron emmitted from the surface has de broglie wavelength lamfha d then?
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