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

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.

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