Question

Explain photoelectric emission on the basis of quantum model of light and derive Einstein's photoelectric emission. Construct logic symbol, Boolean expression and truth table for AND gate.

Answer 1

• A	B	Q
  0	0	0
  0	1	0
  1	0	0
  1	1	1

  TRUTH TABLE (A. B =Q)
• Einstein gave an explanation for photoelectric emission based on the planks quantum theory of light. When light of certain frequency (greater than threshold frequency) incident on a metallic surface the energy on incident photon (hf) is
  

Partly used to eject electron from surfaces.

Partly remains as kinetic energy of ejected electrons.

ie, Energy of photon = work function + kinetic energy of electron

Or hf = $\varphi$+ K. E

Laws of photoelectric emission

There exists a minimum frequency of incident light below which photoelectric emissions is not possible. The frequency is called threshold frequency.

A part of energy of incident electrons in used to eject electrons from a square surface and remaining as kinetic energy of the ejected electrons.

Increase in intensity of the light increase the number of photoelectron.

Increase in frequency of light increase the kinetic energy of electrons.

Derivation of Einstein's photoelectric emission.

Let $m$ be the mass of electron and $V_{max}$ be the maximum velocity of photoelectron by which it will get ejected.

$\frac{1}{2}m{v}_{max}^2=hv-{\varphi }_o$

$\therefore hv={\varphi }_o+\frac{1}{2}m{v}_{max}^2$

But ${\varphi }_o=h{v}_o$

$\therefore hv=h{v}_o+\frac{1}{2}m{v}_{max}^2$

$\therefore \frac{1}{2}m{v}_{max}^2=hv-h{v}_o$

$\therefore \frac{1}{2}m{v}_{max}^2=h(v-v_o)$

This equation is known as Einstein's photoelectric equation.