Modern Physics – answers to sample critical reasoning questions

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Pioneer of Modern Physics - A Einstein

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Modern Physics: the following topics are covered in this page:

Structure of atoms, Bohr’s atom model, X-Rays and Gamma Rays, Radioactivity, Semiconductors, P-N junctions, Diode, Triode,  etc.

Click on a question to see the answer:

1. What is the source of energy released in nuclear fission?

Mass lost in a nuclear fission reaction is the source of energy released.

According to Einstein’s theory of relativity mass and energy are equivalent. They can be converted from one form to the other. If mass m is converted into equivalent amount of energy E, then the relation between them is E=mc^2. Here, c is the speed of light. In a nuclear (fission) reaction the total mass of the products after the reaction is less than the total mass of the reactants before the reaction resulting in a mass loss. The lost mass accounts for the produced energy.

2. Electrons are not there within the nucleus. Then how the \beta-particles are emitted from a radioactive atom?

Nucleus contains only protons and neutrons and not electrons. Yet electrons (\beta-particle) are emitted from the radioactive atoms. Actually, during a \beta-decay a neutron converts permanently into a proton producing one electron which is emits as a \beta-particle.

3. What are the differences between a chemical change and a radioactive change?

In case of a chemical change the the orbital electrons take part in the reactions while the nucleus remains unchanged. On the other hand, in a radioactive reaction change takes place inside the nucleus.
Chemical change does not produce any new element while new elements result from a nuclear change.

4. Effective radius of helium atom is smaller than that of a hydrogen atom – why?

The effective radius of an atom is the radius of the outermost orbit of the electrons and the expression for the radius (r_n) of the n-th orbits is

    \[r_n=\frac{n^2h^2}{4 \pi mZe^2}.\]

Here, Z= mass number = number of protons in the nucleus. For both hydrogen and helium, 1st orbit is the outermost orbit and hence n=1. But for hydrogen Z=1 while for helium Z=2. Therefore from above expression we can conclude that the effective radius of helium is half the effective radius of hydrogen.

5. It is possible that in case of radioactive decay an \alpha-particle emission is followed by two \beta-emissions. Show that the final nucleus is an isotope of the original one.

With the emission of one \alpha particle from the nucleus its atomic number decreases by 2 and when a \beta particle emits the atomic number increases by 1. So if one \alpha and two \beta particles are emitted the atomic number of the final nucleus remains unaltered. Only a decrease in the mass number by 4 is observed. Atomic number remaining same, this results in the production of an isotope.

6. The electron in hydrogen atom is initially in the third excited state. Determine the maximum number of spectral lines which can result when the electron finally moves to the ground state. 

For the third excited state, n=4 and for the ground state n=1. Hence the possible transitions are:

From n_i = 4 to n_f = 3, 2, 1; from n_i = 3 to n_f = 2, 1; and from n_i = 2 to n_f = 1. So, altogether 6 transitions are possible. Therefore, the maximum number of spectral line will be 6.

7. How is nature of a plot that shows the variation of the radius of the orbit (of the electrons) with its principle quantum number.

The expression for the radius (r) in terms of its principle quantum number (n) is:  

    \[r_n=\frac{n^2h^2}{4 \pi mZe^2}.\]

As all the quantities other than r_n and n are constants, we can get r_nn^2. This indicates that the nature of the r_n-n plot is parabolic.

 

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