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Modern Physics - An Important Concept of JEE

Modern Physics - An Important Concept of JEE

Let us first understand what is modern physics? Modern Physics is a branch of physics that deals with the fundamental nature of the universe with post-Newtonian concepts. Galileo Galilei, who is also known as the father of modern physics, contributed significantly in modern physics. In the early twentieth century, some experimental results could not be matched with the predictions of classical physics, which describes physical phenomena at an ordinary scale. Modern physics gradually took birth from these theories.


In JEE, this unit is the most important unit of physics. This unit contains three chapters from the class 12 syllabus, namely, Dual Nature of Matter, Atoms and Nuclei. In this article we will cover the important topics of all the three chapters and understand the elements of modern physics. 


In this article, we will be focusing on the concepts of Modern Physics that play a vital role in the JEE exam.


Important Topics of Modern Physics

  • Photoelectric effect

  • Cathode rays

  • De Broglie equation

  • Rutherford’s Model of Atom

  • Bohr Model

  • Hydrogen Spectrum

  • Composition and Size of Nucleus

  • Radioactivity

  • Nuclear Fission and Fusion.


Important Concepts of Modern Physics

We know that in a chapter there will be many concepts that are being mentioned in the textbook, but when it comes to JEE preparation we have to focus on what is most important among all the concepts in the chapter. Below is the table provided for some most important concepts of Modern Physics.


Sl. No

Name of the Concept

Key Points of Concept


Photoelectric effect

  • The phenomenon of the photoelectric effect revolves around electron emission.

  • The photoelectric effect shows that when light radiation is incident on a metallic surface, the electrons near the surface absorb the radiation energy and get liberated from the surface.

  • These photo or light generated electrons are known as photoelectrons.


Relation between Potential and Photoelectric Current

  • With an increase in the accelerating potential, the photoelectric current increases equally.

  • The saturation current occurs when the photoelectric current reaches its maximum value.

  • All photoelectrons are emitted from the emitter and reach the collection plate when the photoelectrons reach saturation.


Relation between Intensity of the Incident Light and Photoelectric Current

  • Photocurrent does not occur if the frequency of the incident radiation or light is lower than the metal surface's minimum frequency. It rapidly declines until it reaches zero, which is also known as the cut-off frequency or the metal plate's stopping potential.

  • The photoelectric current's cut-off frequency, or stopping potential, is essentially a physical feature of the metal plate.

  • The photocurrent becomes zero or reaches the cut-off frequency only when the stopping potential is sufficient to drive back even the highest-energy photoelectric particles approaching with maximum kinetic energy.


De Broglie hypothesis- dual nature of light

  • De Broglie theory suggests that light behaves like both particles and waves. 

  • According to De Broglie's hypothesis, the wavelength associated with momentum is too small, which explains why microscopic elements do not exhibit wave characteristics.


Rutherfor’s atomic model and its limitation

  • Every atom consists of a tiny central core, called the atomic nucleus in which the entire positive charge and almost entire mass of the atom are concentrated.

  • The atom as a whole is electrically neutral and electrons revolve around the nucleus in various circular orbits.

  • As the electron revolves so it must lose energy continuously, it must spiral inwards and eventually fall into the nucleus. Therefore, atoms should emit a continuous spectrum but what we observe is a line spectrum. So this point isn’t explained by the Rutherford model which is the biggest limitation of this model.


Bohr’s model of hydrogen atom

  • Bohr gives three postulates;

  1. Every atom consists of a central core called nucleus which contains the positive charge and electrons revolve around it in their circular orbits.

  2. Electrons can revolve only in certain discrete non radiating orbits, called stationary orbits for which total angular momentum of the revolving electron is an integral multiple of $\dfrac{h}{2\pi}$, where $h$ is the Planck’s constant.

  3. The emission or absorption of energy occurs only when an electron jumps from one of its specified orbit to another. The difference in the total energy of electrons in the two permitted orbits is absorbed when the electron jumps from inner to the outer orbit.


7.   

Radioactivity

  • Radioactivity is the process by virtue of which a heavy element disintegrates itself without being forced by any external agent to do so.

  • According to laws of radioactive disintegration, the number of atoms disintegrating per second at any instant is directly proportional to the number of radioactive atoms actually present in the sample at that instant.

$-\dfrac{dN}{dt}\varpropto N$

  • The half life of a radioactive element is the amount of time it takes for half of the atoms present in the sample to decay.

  • Average life time of the element is obtained by calculating the total life time of all the atoms of the element and dividing it by the total number of atoms present initially in the sample of the element

8.

Alpha, beta and Gamma decay

  • Alpha decay is the phenomenon of emission of an alpha particle from a radioactive nucleus. Ex:

$U^{238}_{92}\longrightarrow Th^{234}_{90}+He^4_2$

  • Beta decay  is the phenomenon of emission of an electron from a radioactive nucleus. Ex:

$Th^{234}_{90}\longrightarrow Pa^{234}_{91}+e^0_{-1}$

  • Gamma decay is the phenomenon of emission of gamma ray photons from a radioactive nucleus. Ex:

$Ni^{60*}_{28}\longrightarrow Ni^{60}_{27}+E_{\gamma}$

9. 

Nuclear fission and fusion

  • Nuclear fission is the process of splitting a heavy nucleus (usually A>230) into two or more lighter nuclei.

  • Nuclear fusion is the process by which two or more lighter nuclei combine to produce a single heavy nucleus. 


List of Important Formulae of Modern Physics

Sl. No

Name of Concept

Formula

1.

De Broglie wavelength

$\lambda=\dfrac {h}{p}$

Or

$\lambda=\dfrac {h}{\sqrt {2mE}}$

2.

Photon energy

$E=h\nu=\dfrac {hc}{\lambda}$

3.

Einstein’s photoelectric equation and Work function

${E_k}=h\nu-\phi$

Where,

$\phi$- The work function

4. 

Threshold frequency

${\nu_o}=\dfrac {\phi}{h}$

5.

Momentum of photon

The energy of photon moving with momentum p is $E=p c$ 

Therefore, momentum p is given by:

$p=\dfrac {h \nu}{c}$ 

6. 

Bohr’s model of hydrogen atom

From 1st postulate of Bohr’s model of atom we can write;

$\dfrac{mv^2}{r}=\dfrac{KZe^2}{r^2}$

Here, $v$= velocity with which electron is moving and $r$= radius of the orbit.


From 2nd postulate,

$mvr=\dfrac{nh}{2\pi}$


From 3rd postulate,

$E_2-E_1=nh\nu$


The electron's frequency in Bohr's stationary orbit is,

$\nu=\dfrac{KZe^2}{nhr}$


Total energy of electron in Bohr’s stationary orbit is,

$E=-\dfrac{13.6}{n^2}eV$

7. 

Radioactivity

According to the law of radioactive decay, 

$-\dfrac{dN}{dt}\varpropto N$

$R=-\dfrac{dN}{dt}= \lambda N$

Here, $\lambda$ =disintegration                          constant.


Half life ($T_{1/2}$)of radioactive element is,

$T_{1/2}=\dfrac{0.693}{\lambda}$

Average life ($\tau$) is,

$\tau= \dfrac{1}{\lambda}$


The activity of an element is given as, 

$A=A_o e^{-\lambda t}$

            


Solved Examples of Modern Physics

1. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work-function for sodium is 1.82 eV.Find 

(a) the energy of the photons causing the photoelectrons emission. 

(b) the quantum numbers of the two levels involved in the emission of these photons. 

(c) the change in the angular momentum of the electron in the hydrogen atom, in the above transition (Ionization potential of hydrogen is 13.6 eV.)

Sol:

Given, 

Maximum Kinetic Energy for Sodium (KEmax) = 0.73 eV, 

Work-function for sodium (W) = 1.82 eV, 

(a) The energy of photons:

From Einstein’s equation of photoelectric effect,

Energy of photons (E) causing the photoelectric emission = Maximum kinetic energy of emitted photons (KEmax) + work-function (W)

E = KEmax + W = 0.73 eV + 1.82 eV = 2.55 eV, 


(b) For Hydrogen atom, 

E1 = - 13.6 eV, E2 = 3.4 eV, E3 = -1.5 eV, E4 = - 0.85 eV

Now, from the levels given above, E2 - E4 = 2.55 eV

Therefore, quantum numbers of the two levels involved in the emission of these photons are 4 and 2. 


(c) Change in angular momentum in transition from 4 to 2 will be

Here, angular momentum (L) is given by $\dfrac{nh}{2\pi}$, where, n = quantum number and h = planck’s constant, 

Therefore, $\Delta L=L_2-L_4 =  \dfrac{2h}{2\pi}-\dfrac{4h}{2\pi}$

Hence, Change in angular momentum will be $-\dfrac{h}{\pi}$

Key Point: Kinetic energy of the fastest moving electron is maximum kinetic energy only because kinetic energy depends on mass and velocity only. 


2. The mean lives of an unstable nucleus in two different decay processes are 1620 yr and 405 yr, respectively. Find out the time during which three-fourth of a sample will decay.

Sol:

Given, 

Mean life of Process- 1 (t1) = 1620 yr, 

Mean life of Process- 2 (t2) = 405 yr,

Let decay constants of process-1 and process- 2 be λ1 and λ2, respectively. Then, 

$\lambda_1 = \dfrac{1}{t_1}$ 

$\lambda_2 = \dfrac{1}{t_2}$

If the effective decay constant is λ, then

λN = λ1N + λ2

λ = λ1 + λ2

$\lambda = \dfrac{1}{t_1}+\dfrac{1}{t_2}$

$\lambda = \dfrac{1}{1620}+\dfrac{1}{405}$ year-1

$\lambda = \dfrac{1}{324}$ year-1

Now, when three- fourth of the sample will decay. The sample remaining will be one- fourth of the total sample. If the initial sample was N0, then final sample will have N0/4, 

So, $\dfrac{N_0}{4} = N_0e^{-\lambda t}$

Applying logarithms on both the sides, we get, 

$-\lambda t = \ln\lgroup\dfrac{1}{4}\rgroup$ = -1.386

t = 449 yr

Therefore, the time during which three-fourth of a sample will decay will be 449 yr. 

Key point: We need to modify the formula according to the given data. 


Previous Year Questions From Modern Physics

1. The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times {10^{-3}}$ watt will be: (JEE 2021)

  1. ${10^{16}}$

  2. ${10^{15}}$

  3. ${10^{18}}$

  4. ${10^{17}}$

Sol:

Given,

Wavelength of source of monochromatic light=$\lambda$= 600 nm

Power radiated by the source=$3.3 \times {10^{-3}}$ watt

The energy of the photon=E=$\dfrac {hc}{\lambda}= 3.3 \times {10^{-19}} Joules$

Let n is the number of photons emitted per second such that the energy emitted in one second is $3.3 \times {10^{-3}}$ J.

Thus, we get:

nE=$3.3 \times {10^{-3}}$

$n=\dfrac {3.3 \times {10^{-3}}}{3.3 \times {10^{-19}}}={10^{16}}$

Therefore, option A is the right answer.

Trick: Use the formula $P=\dfrac {nhc}{\lambda}$ to arrive at the solution easily.


2. There are $10^{10}$ radioactive nuclei in a given radioactive element, its half-life time is 1 minute. How many nuclei will remain after 30 seconds? (Take, $\sqrt{2}=1.414$) (JEE 2021)

a. $7 \times 10^{9}$

b. $2 \times 10^{9}$

c. $4 \times 10^{10}$

d. $10^{5}$

Sol:

Given that, 

Original number of nuclei, ${N_o} = 10^{10}$

Half- life time, $t_{½} = 60\,s$

To find: Number of nuclei remain after time ($t$ = 30 seconds) i.e, $N$.

To solve this problem we have to use the concept of half life and also the relation of the number of nuclei decayed to the original number of nuclei present in the sample of a radioactive element.

Now using the concept of half live, we can write the relation between the number of nuclei decayed ($N$) to the original number of nuclei($N_o$) as,

$\dfrac{N}{N_o}=(\dfrac{1}{2})^{t/t_{1/2}}$

Now after putting the values of the quantities in the above relation, we get;

$\dfrac{N}{10^{10}}=(\dfrac{1}{2})^{30/60}$

$\dfrac{N}{10^{10}}=(\dfrac{1}{2})^{1/2}$

$N= \dfrac{10^{10}}{\sqrt{2}} \approx 7 \times 10^{9}$

Hence, the number of nuclei that remain after 30 seconds is $7 \times 10^{9}$. 

Therefore, option a is correct.

Trick: The application of the half life concept and the relation between the number of nuclei decayed to the original number of nuclei in a sample is important to solve this problem.


Practice Questions

1. Wavelengths belonging to the Balmer series for hydrogen atoms lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work-function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons.

(Ans: 0.55 eV)

2. A stable nuclei C is formed from two radioactive nuclei A and B with decay constant of 𝜆1 and 𝜆2 respectively. Initially, the number of nuclei of A is N0 and that of B is zero. Nuclei B are produced at a constant rate of P. Find the number of the nuclei of C after time t 

(Ans:$N_c = N_0 (1-e^{-\lambda_1 t})+P \lgroup t+\dfrac{e^{-\lambda_2 t}-1}{\lambda_2}\rgroup$)


Conclusion

In this article we have discussed the Modern Physics chapter with respect to the JEE exam. In this article we have provided solved examples, previous year questions, important formulae list, etc..

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1. How many questions come from Modern Physics in JEE?

Modern Physics is having high importance in the JEE exam, it carries nearly 3-4 questions with a weightage 11%.

2. Which chapters are a part of Modern Physics?

This unit contains three chapters from the class 12 syllabus, namely, Dual Nature of Matter, Atoms and Nuclei.

3. Which chapter has more weightage in JEE?

Modern Physics, from the class 12th syllabus, is the most important topic for JEE. Current Electricity & Magnetism, Electrostatics and Optics are also very important chapters for JEE.