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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.

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.

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.

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 + λ2N

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.

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)

${10^{16}}$

${10^{15}}$

${10^{18}}$

${10^{17}}$

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.

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

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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FAQ

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.

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