Question: Q. 6. Using photon picture of light, show how Einstein’s photoelectric equation can be established. Write two features of photoelectric effect which cannot be explained by wave theory. U [OD I 2017]

ns. In the photon picture, energy of the light is assumed to be in the form of photons, each carrying an energy $hv$.

Einstein assumed that photoelectric emission occurs because of a single collision of a photon with a free electron.

The energy of the photon is used to

(i) Free the electrons from the metal.

[For this, a minimum energy, called the work function $(={\varphi }_0)$ is needed].

and

(ii) Provide kinetic energy to the emitted electrons. $1/2$ Hence,

or

$$ \begin{aligned} & (K . E .){\max }=h v-\phi{0} \ & \left(\frac{1}{2} m v_{\max }^{2}=h v-\phi_{0}\right) \end{aligned} $$

This is Einstein’s photoelectric equation $1/2$ Two features (which cannot be explained by wave theory) :

(a) ‘Instantaneous’ emission of photoelectrons

(b) Existence of a threshold frequency

(c) ‘Maximum kinetic energy’ of the emitted photoelectrons, is independent of the intensity of incident light

$$1/2+1/2$$

(Any two)

[CBSE Marking Scheme 2017]

Q. 7. (i) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?

(ii) The work function of the following metals is given : $\mathrm{Na}=2.75\mathrm{eV},\mathrm{K}=2.3\mathrm{eV},\mathrm{Mo}=4.17\mathrm{eV}$ and $\mathrm{Ni}=5.15\mathrm{eV}$. Which of these metal will not cause photoelectric emission for radiation of wavelength $3300\text{AA}$ from a laser source placed $1\mathrm{m}$ away from these metals? What happens if the laser source is brought nearer and placed $50\mathrm{cm}$ away.

U] [Delhi I 2017]