Question
Write Einstein's photoelectric equation and explain any two observations related to the photoelectric effect.
✓
Answer
The photoelectric equation of Einstein,
$\frac{1}{2} mv _{\max }^2= hv -\phi_0$
Observations related to photoelectric effect:
(1) On frequency of incident light is less than the threshold frequency or $v<v_0$ then from equation $\frac{1}{2} \operatorname{mv}_{\max }^2= h \left(v-v_0\right)$, the value of kinetic energy of electron will be negative since kinetic energy is negative. So if $v<v_0$ emission of photo-electrons is not possible, i.e. the frequency of incident light must be greater than or equal to the threshold frequency.
(2) The number of electrons emitted depends on the number of incident photons. In other words, we can say that the number of electrons emitted in unit time for light of fixed frequency from unit area of surface, depends on the number of photons incident on unit area in unit time on the surface i.e. the intensity of light.
(3) From eq. $\frac{1}{2} m v_{\max }^2=h\left(v-v_0\right)$ it is clear that kinetic energy of emitted photoelectrons depends on the frequency of the incident proton but not on the intensity of light.
(4) According to quantum theory, energy of light remains in the form of photons. Therefore as soon as a photon of sufficient energy strikes on the surface, an electron is emitted due to the absorptions, i.e. there is no time lag in the emission of photoelectrons.
$\frac{1}{2} mv _{\max }^2= hv -\phi_0$
Observations related to photoelectric effect:
(1) On frequency of incident light is less than the threshold frequency or $v<v_0$ then from equation $\frac{1}{2} \operatorname{mv}_{\max }^2= h \left(v-v_0\right)$, the value of kinetic energy of electron will be negative since kinetic energy is negative. So if $v<v_0$ emission of photo-electrons is not possible, i.e. the frequency of incident light must be greater than or equal to the threshold frequency.
(2) The number of electrons emitted depends on the number of incident photons. In other words, we can say that the number of electrons emitted in unit time for light of fixed frequency from unit area of surface, depends on the number of photons incident on unit area in unit time on the surface i.e. the intensity of light.
(3) From eq. $\frac{1}{2} m v_{\max }^2=h\left(v-v_0\right)$ it is clear that kinetic energy of emitted photoelectrons depends on the frequency of the incident proton but not on the intensity of light.
(4) According to quantum theory, energy of light remains in the form of photons. Therefore as soon as a photon of sufficient energy strikes on the surface, an electron is emitted due to the absorptions, i.e. there is no time lag in the emission of photoelectrons.
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