WBCHSE Class 12 Physics Atom Short Question And Answers

Atom Short Question And Answers

Question 1. The Electron of a hydrogen atom revolves In the 3rd Bohr orbit. Find the angular momentum of the electron. (h = 6.6 × 10 -27 erg .s)
Answer:

Principal quantum number of 3rd Bohr orbit, n = 3. So according to Bohr’s postulate, the angular momentum of an electron in that orbit,

L3 = \(3 \times \frac{h}{2 \pi}=\frac{3 \times 6.6 \times 10^{-27}}{2 \times \pi}\)

3.15 × 10 -27erg . s

Question 2. In a hydrogen atom, how many times is the radius of the fourth orbit compared to that in the second orbit?

If the principal quantum number of an orbit is n, then radius rn ∝= n²

∴  \(\frac{\text { radius of fourth orbit }}{\text { radius of second orbit }}=\frac{(4)^2}{(2)^2}\)

= \(\frac{16}{4}\)

= 4

Therefore, the radius of the fourth orbit is 4 times the radius of the second orbit

Question 3. In a hydrogen atom, how many times is the speed of the electron in the fourth orbit compared to that in the second orbit?
Answer:

If the principal quantum number of an orbit is n, then the speed ν ∝ 1/n

So \(\frac{\text { speed of electron in the fourth orbit }}{\text { speed of electron in the second orbit }}=\frac{1 / 4}{1 / 2}\)

= \(\frac{1}{2}\)

Therefore, the speed of an electron in the fourth orbit is half of that in the second orbit

Question 4. The electron in a hydrogen atom is excited to the n-th © excited state. How many possible spectral lines can it emit In transition to the ground state?
Answer:

There are n number of states from the n-th quantum

State to the ground state (n = 1 ). The transition may occur between any pair of these states.

Since for each transition, a spectral line is formed, the number of possible spectral lines = \({ }^n C_2=\frac{n(n-1)}{1 \times 2}=\frac{1}{2} n(n-1)\)

Question 5. explain why the spectrum of hydrogen contains several lines although a hydrogen atom has only one electron
Answer:

In hydrogen gas, there are innumerable hydrogen atoms, i.e., in a hydrogen discharge tube, there are innumerable electrons. Different electrons absorb different amounts of energy and get excited In different states. As a result, when they radiate energy, transitions occur between different pairs of states; for each transition, a spectral line Is obtained. Thus, the spectrum of hydrogen contains several lines

Question 6. Write two important limitations of Rutherford’s nuclear
Answer:

Two important limitations of Rutherford’s nuclear model of the atom are

  1. This model cannot explain the stability of matter.
  2. It cannot explain the characteristic line spectra of atoms of different elements

Question 7. What is the significance of the negative energy of an electron in its orbit?
Answer:

The energy of an electron in its atomic orbit is negative. Due to the nuclear attraction on it. It may come out of an the atom to become a zero-energy free electron only if an equal amount of = -13.6 eV positive energy is supplied to it from outside

Question 8. What is meant by excitation potential and ionization Potential
Answer:

The excitation potential is the potential to be applied to an electron in the outermost shell of an atom to lift it to an excited energy state. On the other hand, the ionization potential is the minimum potential to be applied to that electron to bring it outside the atom so that the atom is ionized. For example, the excitation potential of the ground state electron of a hydrogen atom is 10.2 V for transition to the first excited state, whereas the ionization potential of that electron is 13.6 V

Question 9. Find the wavelength of electromagnetic waves of frequency 5 × 10-19Hz in free space. Where is this type of wave used?
Answer:

λ = \(\left(\frac{c}{v}=\frac{3 \times 10^8}{5 \times 10^{19}}=6 \times 10^{-12} \mathrm{~m}\right)\)

= 6 ×10 -12× 10-10

= 0.06 Å

It is hard X-ray-used in the treatment of cancer-affected cells, or the detection of internal defects of metals.

Question 10. Show that the energy of the first excited state of He+ is equal to the speed of the electron In the first orbit of the hydrogen atom
Answer:

En = \(-\frac{m Z^2 e^4}{2 n^2 \hbar^2}\)

∴ \(\left(\frac{Z_1}{Z_2}\right)^2 \cdot\left(\frac{n_2}{n_1}\right)^2\)

= \(\left(\frac{2}{1}\right)^2 \times\left(\frac{1}{2}\right)^2\)

Or , E1 = E2

Question 11. Show that the speed of an electron in the second orbit of the He+ atom is equal to the speed of an electron in the first orbit of the hydrogen atom.
Answer:

We have \(\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{n \hbar}\)

⇒ \(\frac{v_1}{v_2}=\frac{Z_1}{Z_2} \cdot \frac{n_2}{n_1}\)

= \(\frac{2}{1} \times \frac{1}{2}\)

= 1

Or, v1 = v2

Question 12. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer: 

Ground state energy of hydrogen at room temperature = 13.66 eV. The energy obtained after bombardment by the 12.5 eV electron beam is (-13.6+12.5) =-1.1 eV which lies between the third (-1.51 eV) and fourth (-0.85 eV) energy levels of the hydrogen atom. Thus the wavelength ofthe emitted radiations may lie in the Lyman ( 103 nm and 122 nm) or Balmer (656 nm ) series.

Question 13. Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n- 1 ). For large n, show that this frequency equals the classical frequency of revolution ofthe electron in the orbit.

The energy of an electron in the n-th energy level of the hydrogen atom

⇒ \(E_n=\frac{-2 \pi^2 m e^4}{\left(4 \pi \epsilon_0\right)^2 \cdot n^2 h^2}\)

⇒  \(E_n-E_{n-1}=\frac{2 \pi^2 m e^4}{\left(4 \pi \epsilon_0\right)^2 \cdot h^2} \cdot \frac{2 n-1}{n^2(n-1)^2}\)

⇒  \(f=\frac{E_n-E_{n-1}}{h}\)

= \(\frac{2 \pi^2 m e^4}{\left(4 \pi \epsilon_0\right)^2 h^3} \cdot \frac{2 n-1}{n^2(n-1)^2}\)

Question 14. The total energy of an electron in the first excited state ofthe hydrogen atom is about -3.4 eV. In this state

  1. What is the kinetic energy ofthe electron?
  2. What is the potential energy of the electron?

Answer: 

According to Bohr’s theory

1. The kinetic energy of the electron

= – Total energy ofthe electron

= 3.4 eV

2. Potential energy = -2K.E.

= -6.8 eV

Question 15. If Bohr’s quantization postulate (angular momentum = nh/2π ) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak ofquantisation of planets around the sun?

In the case of planetary motion, the value of angular momentum is much larger than the value of h (In the case of Earth its value is 1070h ). As a result, the value of n becomes nearly 1070h, making the difference between the successive energy levels infinitesimally small and the levels may be considered continuous.

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