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Light quanta with an energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV.find the maximum impulse transmitted to the surface of the metal when each electron flies out.?
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Light quanta with an energy 4.9 eV eject photoelectrons from metal wit...
Explanation:


Work Function:


The minimum energy required to eject an electron from the surface of a metal is called the work function. It can be denoted by Φ.

Energy of Light Quanta:


The energy of a light quantum is given by the equation:

E = hf

where E is the energy of the light quantum, h is Planck's constant, and f is the frequency of the light.

Impulse Transmitted to the Surface:


When a photoelectron is ejected from the surface of a metal, it carries some momentum with it. This momentum is transferred to the surface of the metal, resulting in an impulse. The maximum impulse is obtained when the photoelectron is ejected at an angle of 90° to the surface of the metal.

Calculation:


Given, energy of light quanta, E = 4.9 eV
Work function, Φ = 4.5 eV

The kinetic energy of the ejected photoelectron can be calculated as:

KE = E - Φ
KE = 4.9 - 4.5
KE = 0.4 eV

The momentum of the photoelectron can be calculated as:

p = sqrt(2mKE)

where m is the mass of the electron.

Using the value of m = 9.11 x 10^-31 kg and KE = 0.4 eV, we get:

p = 1.51 x 10^-25 kg m/s

The maximum impulse transmitted to the surface can be calculated as:

I = 2p
I = 2 x 1.51 x 10^-25
I = 3.02 x 10^-25 Ns

Therefore, the maximum impulse transmitted to the surface of the metal is 3.02 x 10^-25 Ns.
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Light quanta with an energy 4.9 eV eject photoelectrons from metal wit...
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Light quanta with an energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV.find the maximum impulse transmitted to the surface of the metal when each electron flies out.?
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