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 $h v$.
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 $\left(=\phi_{0}\right)$ 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 $3300 \AA$ 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]
Show Answer
Solution:
Ans. (i) Einstein’s Photoelectric equation is
$$ h v=\phi_{0}+K_{\max } $$
When a photon of energy ’ $h v$ ’ is incident on the metal, some part of this energy is utilized as work function to eject the electron and remaining energy appears as the kinetic energy of the emitted electron.
(ii) $E=\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3.3 \times 10^{-7} \times 1.6 \times 10^{-19}}$ $=3.77 \mathrm{eV}$
The work function of $\mathrm{Mo}$ and $\mathrm{Ni}$ is more than the energy of the incident photons, so photoelectric emission will not take place from these metals. $1 / 2$ Kinetic energy of photoelectrons will not change, only photoelectric current will change.
[CBSE Marking Scheme 2017]
