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.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The plates of a capacitor are 2.00cm apart. An electronproton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released?
A charged particle oscillates about its mean equilibrium position with a frequency of $10^9 $ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
→Consider the situation of the previous problem. Consider the fastest electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.
A capacitor of capacitance C is given a charge Q. At t = 0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the second capacitor as a function of time.
Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at,
- The mid-point of a side,
- At the centre of the triangle.
A long solenoid with 15 turns per cm has a small loop of area $2.0 cm^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing?
→A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is $\mu.$ The block is given an initial speed $\nu_0.$ As a function of the speed $\nu$ write,
→The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge $+q_1, -q_2$ is modified to
$|\text{F}|=\frac{\text{q}_1\text{q}_2}{(4\pi\in_0)}\frac{1}{\text{r}^2}.\text{r}\geq\text{R}_0$
$=\frac{\text{q}_1\text{q}_2}{(4\pi\in_0)}\frac{1}{\text{R}^2_0}\Big(\frac{\text{R}_0}{\text{r}}\Big)^{\in}.\text{r}\geq\text{R}_0$
Calculate in such a case, the ground state energy of a H-atom, if $\in=0.1,\text{R}_0=1\mathring{\text{A}}$.
→$|\text{F}|=\frac{\text{q}_1\text{q}_2}{(4\pi\in_0)}\frac{1}{\text{r}^2}.\text{r}\geq\text{R}_0$
$=\frac{\text{q}_1\text{q}_2}{(4\pi\in_0)}\frac{1}{\text{R}^2_0}\Big(\frac{\text{R}_0}{\text{r}}\Big)^{\in}.\text{r}\geq\text{R}_0$
Calculate in such a case, the ground state energy of a H-atom, if $\in=0.1,\text{R}_0=1\mathring{\text{A}}$.
The switch S shown in figure. is kept closed for a long time and is then opened at t = 0. Find the current in the middle $1.0\Omega$ resistor at t = 1ms.
→Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2\lambda$. Consider a line passing through $S_2$ and perpendicular to the line $S_1S_2$. What is the smallest distance from $S_2$ where a minimum of intensity occurs?
→