GRADE |12 |PHYSICS | PHOTON| IMPORTANT IMPORTANT SHORT QUESTION AND ANSWER

 

Question Answer

Short discussion questions

1.

The quantum of radiation is called a photon. Its characteristics are:

(i) The momentum of a photon is given by p =  hλ$\frac{\mathrm{h}}{\lambda }$

(ii) The rest mass of photon is zero. Hence, a photon cannot be at rest.

 

2.

 The electrons obtained from the photoelectric effects are called photoelectrons. No they are not different from the ordinary electrons.

3.

A substance emits photoelectron when the energy to the incident photon is greater than the work function of the substance. To emit photoelectron from a substance, the energy of the incident light should be more than the work function of the substance. Since, the energy of the light from the bulb is less than the work function of wood, so no photo electrons are emitted from a wooden table.

 

4.

Since K.E = hf - θ ….1

If the frequency is doubled

K.E’ = 2hf - θ …..2

 

Dividing 2  by 1we get  

K.E′K.E=2hf−θhf−θ$\frac{\mathrm{K}.{\mathrm{E}}^{\prime }}{\mathrm{K}.\mathrm{E}}=\frac{2\mathrm{h}\mathrm{f}-\theta }{\mathrm{h}\mathrm{f}-\theta }$

K.E′K.E=2hf−2θ+θhf−θ=2+θhf−θ$\frac{\mathrm{K}.{\mathrm{E}}^{\prime }}{\mathrm{K}.\mathrm{E}}=\frac{2\mathrm{h}\mathrm{f}-2\theta +\theta }{\mathrm{h}\mathrm{f}-\theta }=2+\frac{\theta }{\mathrm{h}\mathrm{f}-\mathrm{\theta }}$which is greater than 2

Hence the kinetic energy becomes slightly more than double.

 

5.

It is because the number of the emitted photoelectron is independent to the frequency of incident radiation.

 

6.

Photoelectric emission	Thermionic emission
It is the emission of the photoelectrons When light of the suitable frequency falls on the metal	It is the emission of the electrons from the surface of hot metal source.
This effect is temperature independent.	This effect is temperature dependent

 

7.

8.          

Work function of copper ϕ1 = 4.5ev

Work function of sodium ϕ2 = 2.0ev

Energy of the radiation = hcλ$\frac{\mathrm{h}\mathrm{c}}{\lambda }$  = 6.626∗10−34∗3∗1084000∗10−10$6.626\ast \frac{{10}^{-34}\ast 3\ast {10}^8}{4000\ast {10}^{-10}}$  = 4.9 *10-19 J= 3.09ev

Since the energy of radiation is very less than the work function of the copper photoelectric emission doesn’t take place while in the case of sodium work function of the metal is less than energy of the incident radiation photoelectric emission takes place.

 

9.

The applications of photoelectric effect are:

(i) Photoemissive cell

(ii) Photovoltaic cell

(iii) Photoconductive cell.

 

10.

 

Long discussion questions

1.

The election of electrons from metallic surface when light is incident on it is known as photoelectric effect.

According to Einstein’s (in 1905) quantum theory of radiation, light is a particle called quantum and the energy carried by each quantum s called photon. The rest mass energy of photon is zero. THe energy of photon having frequency ‘f’ is given by:

E = hf = hcλ$\frac{\mathrm{h}\mathrm{c}}{\lambda }$

Where, h = Plank’s constant

C = speed of light.

If light energy (photon) falls on any surface, it is used up in two ways by the surface.

(i) First part of energy called minimum energy is used to excite the electron in the atom and brings to the surface. This energy is called threshold energy; denoted by θ and given as:

Or, θ = f=hfo = hcλo$\frac{\mathrm{h}\mathrm{c}}{{\lambda }_{\mathrm{o}}}$

Where, f = threshold frequency of light.

(ii) The remaining part of light energy provides the K.E. of the emitted photoelectrons.

Ie. K.Emax = 12$\frac{1}{2}$mv2max

Where, m = mass of emitted photoelectron and,

Vmax = Maximum speed of emitted electron.

Therefore, we can write.

E = θ + K.E.max

Or, hf = hf0 + 12$\frac{1}{2}$mv2max

Or, h (f – fo) = 12$\frac{1}{2}$mv2max

This equation (i) is called Einstein’s photoelectric equation.

 

2.

Introduction:

An experiment set – up by Milikan to determine the value of Plank’s constant and hence to verify the Einstein’s photoelectric effect (equation) is called Milikan’s photoelectric experiment.

Experimental setup: The experimental set up for this experiment is shown as below.

Description:

It consists of a glass tube at the center which a rotating wheel W is present in which alkali metal (like Li, Na, K etc.) are coated. These alkali metal acts as cathode and cup shaped anode is present. These electrodes (anode and cathode) are connected through a variable battery and electrometers (ammeter). A knife is used to scratch and remove the metal oxide from cathode (if formed). There is a window through which light passes the tube.

Working:

Light is allowed to fall of cathode. Due to the photoelectrons are emitted and move towards anode (being the potential). This is observed as photon current in electrometer ‘E’. Then the negative potential in begin to increases at anode ‘A’, this results the decrease in photocurrent for a particular value of – ve potential is called stopping potential denoted by Vs.

Therefore, we can write,

eVs = 12$\frac{1}{2}$mv2max = K.E.max…(i)

From einstein’s photoelectric equation, we write,

E = θ + K.E.max

Or, hf = hfo + eVs

Or, Vs = he$\frac{\mathrm{h}}{\mathrm{e}}$f - he$\frac{\mathrm{h}}{\mathrm{e}}$fo …(ii)

This, equation (ii) is a straight line of the form y = mk + c, where slope m = he$\frac{\mathrm{h}}{\mathrm{e}}$ and intercept c = −hfoe$-\frac{\mathrm{h}{\mathrm{f}}_{\mathrm{o}}}{\mathrm{e}}$ as shown in figure,

Now, perform the experiment with different frequency of light and corresponding stopping potential are observed. If the plots of Vs versus f is also of the form y = mx + c, then Einstein’s photoelectric equation is said to be verified.

Let, the slope of experimental result from graph be:

m = ba$\frac{\mathrm{b}}{\mathrm{a}}$ = changeinstoppingpotentialchangeinfrquency$\frac{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{p}\mathrm{p}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{f}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}}$

From equation (ii), we have,

Slope = he$\frac{\mathrm{h}}{\mathrm{e}}$

Or, ba=he$\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{h}}{\mathrm{e}}$

Or, h = ba$\frac{\mathrm{b}}{\mathrm{a}}$ * e.

Where, ‘e’ is the electronic charge.

Thus, knowing the value of a,b and e in above expression the value of Plank’s constant can be determined.

Its value was found to be 6.62 * 10-34J/s.

 

3.

 According to the Einstein, light of frequency   consists of a photons which has the energy equal to hf . When the photon of such light is incident on the metal, its energy is completely transferred to the electron in the metal.  A part of the energy is acquired by the electron to pull out the electron and the rest of the energy is imparted to the emitted electron in the form of K.E energy .theses emitted electron are called photoelectron.

 The Einstein photoelectric equation is given by

E = θ + K.E.max

Or, hf = hfo + eVs

where θ is the work function of the metal.

It is defined as the minimum energy requires just to liberate an electron from the metal surface with zero velocity.

 

Vs

is the stopping potential defined as the minimum negative potential to anode plate for which photoelectric current becomes Zero

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