JEE Important Chapter - Atoms and Nuclei

The chapter Atoms and Nuclei gives us the idea about studying the structure of atoms and nuclei. This chapter is divided into two subcategories, the first one is atoms in which the concepts like Rutherford’s atom model, atomic spectra, Bohr’s model of hydrogen atom and energy level diagram have been discussed. The second one consists of concepts like nuclear size, nuclear density, nuclear binding energy, radioactivity and nuclear fission and fusion.

Now, let's move on to the important concepts and formulae related to JEE and JEE main physics atoms and nuclei exam along with a few solved examples.

JEE Main Physics Chapter-wise Solutions 2022-23 

Important Topics of Atoms and Nuclei

• Rutherford's atomic model and its limitation

• Bohr model of hydrogen atom

• Bohr’s explanation of spectral series of hydrogen atom

• Nuclear size and nuclear density

• Mass energy relation and nuclear binding energy 

• Radioactivity

• Alpha, beta and Gamma decay

• Nuclear fission and fusion

Atoms and Nuclei Important Concept for JEE

List of Important Atoms and Nuclei Formulas 

Difference Table Between Nuclear Fission and Fusion

Solved Examples 

1. Calculate the number of revolutions per second done by a hydrogen atom electron in the third Bohr orbit.

Sol: 

Given that, Bohr orbit ($n$) = 3

We have to calculate the frequency i.e, $\nu$.

In order to calculate the frequency of revolution we have to use the second postulate of Bohr’s atomic model according to which angular momentum of the electron should be equal to the integral multiples of $\dfrac{h}{2 \pi}$.

According to the second postulate of Bohr’s atomic model,

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

As we know that $v=\omega r$, so after putting this in the above equation we get,

$m\omega r^2=\dfrac{nh}{2\pi}$

$m (2\pi \nu) r^2=\dfrac{nh}{2\pi}$...........(As $\omega=2\pi \mu$)

$\nu = \dfrac{nh}{4\pi^2 m r^2}$

Now after putting the values of mass of electron($m=9.1 \times 10^{-31}\,kg$), Planck’s constant($h=6.6 \times 10^{-34}\, J/s$, radius of orbit($r=0.53 \times 10^{-10}\,m$) and $n=3$, we get;

$\nu = \dfrac{3 \times 6.6 \times 10^{-34}}{4 \times (3.14)^2 \times 9.1 \times 10^{-31} (0.53 \times 10^{-10})^2}$

On further simplification we get;

$\nu = \dfrac{19.8 \times 10^{-34} }{100.81 \times 10^{-51}}$

$\nu = 0.1964 \times 10^{17}\,rev/sec$

$\nu = 1.964 \times 10^{16}\,rev/sec$

Hence, the frequency of revolution done by an electron in the third orbit of a hydrogen atom is $1.964 \times 10^{16}\,rev/sec$.

Key point: The knowledge of the second postulate in Bohr’s atomic model is necessary to solve this problem.

2. A radioactive substance's mean life is 1500 years for alpha emission and 300 years for beta emission, respectively. Calculate the time it takes for three-quarters of a sample to decay if it is decaying by both alpha and beta emission at the same time. (Take, $\log_{10}4= 1.386$)

Sol: 

Given that,

Mean life of alpha emission, $\tau_\alpha$ = 1500 years

Mean life of beta emission, $\tau_\beta$ = 300 years

Number of nuclei remains, $N={N_o} - \dfrac{3}{4}{N_o}=\dfrac{N_o}{4}$

To find: Time during which it decay to three fourths of a sample i.e, $t$.

To solve this problem we have to apply the concept of mean life time and also the laws of disintegration of radioactive elements. If we consider the $\lambda_\alpha$ and $\lambda_\beta$ be the decay constants of the alpha and beta emission then using the formula of mean life time, we can write;

$\lambda_\alpha=\dfrac{1}{\tau_\alpha}=\dfrac{1}{1500}$

And, $\lambda_\beta=\dfrac{1}{\tau_\beta}=\dfrac{1}{300}$

Total decay constant is given as, 

$\lambda=\lambda_\alpha+\lambda_\beta$

$\lambda= \dfrac{1}{1500}+\dfrac{1}{300}$

$\lambda=\dfrac{6}{1500}=\dfrac{1}{250}\,yr^{1}$

Now, from laws of disintegration we can write,

$N=N_o e^{-\lambda t}$

After putting the values of $N$ , we get;

$\dfrac{N_o}{4}=N_o e^{-\lambda t}$

$t= \dfrac{\log_{e}4}{\lambda}$

Now after putting the value of $\lambda$, we get;

$t= 2.3026 (\dfrac{\log_{10}4}{1/250})$

$t = 2.3026 \times 1.386 \times 250$

$t = 797.8$ years

Hence, the time taken by the sample to decay its three fourths value both by alpha and beta emission is 797.8 years.

Key point: The concept of mean life time of a radioactive sample and laws of radioactive disintegration is important to solve this problem.

Previous Year Questions from JEE Paper

1. 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 Main 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.

Key point: 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.

2. A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn disintegrates to some other stable nucleus C. Plot of a graph showing the variation of number of atoms of nucleus B versus time is :

(Assume that at t = 0, there are no B atoms in the sample) (JEE Main 2021)

a.

b.

c.

d. 

Sol:

Given that, initially the number of atoms of $B=0$ at $t=0$ and the last product of reaction i.e, nuclei $C$ is a stable one. Now we have to find the behaviour of $B$ with respect to time as it decays to $C$. 

To answer this question we have to use the concept of growth and decay of a radioactive sample. As it is mentioned that the number of $B$ atoms are zero at $t=0$ so the graph must start from the origin. When the rate of decay of B equals the rate of production of B, the number of atoms of B begins to increase and reaches a maximum value. Because growth and decay are both exponential functions, the number of atoms will start to decrease after that maximum value, hence the best possible graph is (d). Therefore, option d is correct.

Key point: The concept of growth and decay of a radioactive sample is essential to solve this problem.

Practice Questions

1. Which level of doubly ionised lithium has the same energy as the ground state energy of a hydrogen atom? It is necessary to compare the orbital radii of the two levels. (Ans: 3,3)

2. The decay constant for the radioactive nuclide $Cu^{64}$ is $1.5 \times 10{-5}\,s{-1}$. Find the activity of a sample containing $1\,\mu g$ of $Cu^{64}$. The atomic weight of copper is equal to 63.5 g/mole. The mass difference between the supplied radioisotope and regular copper is ignored.(Ans: 3.87 Ci)

Conclusion

In this article we have discussed the various models related to the structure of atoms and nuclei like Rutherford’s model of atom and Bohr’s model of atom. Bohr's model of atoms is widely accepted because it explained properly about the line spectra of hydrogen atoms which were not explained by Rutherford's model of atoms. We have also mentioned the various important atoms and nuclei formulas. We have also discussed topics like nuclear size, nuclear density, radioactivity, mass-energy relation and binding energy of nuclei.

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