IIT JEE , AIEEE Forum - Physics , Modern Physics - Hello edison sir : its photo electric effect

gajju

why photoelectric effect does not take place for free electron, i.e. it takes place for bound electron only?

Moreover why how compton scattering is diferent from photo electric effect?

edison

PHOTOELECTRIC EFFECT (PEE)

Generally, when X-ray photon of say 100Kev photon is incident on lithium, NO PEE will take place as electron in Lithium atom is nearly free w.r.t X-ray photon energy so there is no Photo electric absorption probability. On the contrary, if same X-ray photon is made to incident on high Z material say lead, then definitely PEE will take place as the electron is in bound state w.r.t energy of X-ray photon. Similarly, when UV photon is made to incident on Li atom PEE will take place as w.r.t UV photon energy the electron in Lithium is in bound state.

Thus, we conclude that photoelectric effect takes place for bound electron but not for free electrons

REASON: It is because for photo electric effect to take place the CONSERVATION OF MOMENTUM IS MUST. And also conservation of SPIN should be there. So, for bound electron the momentum conservation takes place easily.

But the momentum is not conserved for free electrons resulting into effectively zero cross section for absorption of photon for photo electric effect to take place in case of free electrons.

edison

To Prove: That for free electron no conservation of momentum takes place, thus no photoelectric effect will occur.

Proof: Let us consider a free electron of mass m_e,

Thus force on electron = 0

Or electron is moving with constant initial velocity = V₁

Now when photon of energy = h?, is made to incident on free electron then for photo electric effect to take place the entire photon must be absorbed.

So applying conservation of Momentum before and after photon absorption we have,

h?/c + m_eV₁ = m_eV₂

or V₁ = V₂ - h?/m_ec ?(1)

Also applying conservation of kinetic energy we have

h? + ( ½)m_eV₁² = (½) m_eV₂²

or V₁² = V₂² - 2 h?/m_e ?(2)

Putting (1) in (2) we obtain,

(V₂ - h?/m_ec )² = V₂² - 2 h?/m_e

or V₂² - 2(h?/m_ec) V₂ + (h?/m_ec )² = V₂² - 2 h?/m_e

or - 2(h?/m_ec) V₂ + (h?/m_ec )² = - 2 h?/m_e

or V₂ = [(h?/m_ec )² + 2 h?/m_e] m_ec/2h?

or V₂ = h?/2m_ec + c

thus, V₂ > c, but velocity of electron can never be greater than speed of light so, PEE can not take place for free electron.

Conservation of SPIN NUMBER: Since before PEE for free electron the angular momentum of photon is 1 (in units of h cross) and that of electron is ½

So before PEE total angular momentum = 3/2 (in units of h cross)

After collision = ½ (as there is only electron but no photon)

So PEE can not take place for free electron.

But for bound electron the there is nucleus to conserve angular momentum.

edison

Compton effect is different from PEE in the sense that in PEE entire photon is absorbed whereas in compton scaterring/effect the photon transfers some of its energy to electron and then it is scaterred.

PEE effect does not take place for free electrons, but is for bound electrons only.

Compton effect takes place for free or nearly free electrons only.

Moreover, in Compton effect as the photon is scattered so initially and finally, electron and photon both are present so that the spin is conserved.

gajju

thank you very much sir.

visioniit

thanx sir.

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