Plus Two Physics Chapter Wise Questions and Answers Chapter 12 Atoms

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Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 12 Atoms

Plus Two Physics Atoms NCERT Text Book Questions and Answers

Question 1.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
Since

For Paschen series, n₁ = 3 and n₂ = ∞

Question 2.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and m = 3 orbits?
Answer:
Given r₁ = 5.3 × 10^-11m
n₁ = 1, n₂ = 2, r₂ = ?
n₃ = 3, r₃ = ?
Since r ∝ n²
∴ r₁ ∝ n₁²
And r₂ ∝ n₂²

or r₂ = 4 × r₁ = 4 × 5.3 × 10^-11
or e₂ = 2.12 × 10^-10m
Also r₃ ∝ n₃²

or r₃ = 9r₁ = 9 × 5.3 × 10^-11
or r₃ = 4.77 × 10^-10m.

Question 3.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelength will be emitted?
Answer:
First excitation energy
E = E₂ – E₁ = -3.4 – (-13.6) = 10.2 eV
Second excitation energy
E = E₃ – E₁ = -1.51 -(-13.6) = 12.09 eV
Third excitation energy
E = E₄ – E₁ = -0.85 – (-13.6) = 12.75 eV
Second incident beam has energy = 12.5 eV
So only first two lines in the Lyman series of wave-length 103 nm and 122 nm will be emitted.
Also E₃ – E₂ = -1.51 – (-3.4) = 4.91 eV
i.e, first line in the Balmer series of wavelength 656nm will also be emitted.

Question 4.
In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ ms^-1.
Answer:
Here r= 1.5 × 10¹¹ m,
v = 3 × 10⁴ ms^-1, m = 6 × 10²⁴kg, n = ?

Plus Two Physics Atoms One Mark Questions and Answers

Question 1.
A radioactive nucleus emits beta particle. The parent and daughter nuclei are
(a) isotopes
(b) isotones
(c) isomers
(d) isobars
Answer:
(d) isobars

Question 2.
Consider an electron in the n^th orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength λ of that electron as
(a) (0.529) nλ
(b) \(\sqrt{n} \lambda\)
(c) (13.6)λ
(d) nλ
Answer:
(d) nλ

Question 3.
The ionization energy of hydrogen atom is 13.6 eV. Find the energy corresponding to a transition between 3^rd and 4^th orbit.
Answer:
E = E₄ – E₃

= -0.85 + 1.51 = 0.66 eV.

Question 4.
If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is
(a) 10.2 eV
(b) 0 eV
(c) 3.4 eV
(d) 6.8 eV
Answer:
(c) 3.4 eV
Explanation:

= 3.4 eV.

Question 5.
In the Bohr model of the hydrogen atom, the lowest orbit corresponds to
(a) infinite energy
(b) maximum energy
(c) minimum energy
(d) zero energy
Answer:
(c) minimum energy.

Question 6.
Write down the Balmerformula for wavelength of H_a line.
Answer:

Plus Two Physics Atoms Two Mark Questions and Answers

Question 1.
Given Rydberg constant as 1.097 × 10^-7m^-1. Find the longest and shortest wavelength limit of Baler Series.
Answer:
\(\bar{v}=\frac{1}{\lambda}=R_{H}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
Longest wavelength n₁ = 2 and n₂ = 3

Shortest Wavelength n₁ = 2 and n₂ = α

Plus Two Physics Atoms Three Mark Questions and Answers

Question 1.
Bohr combined classical and early quantum concept and gave his theory in the form of three postulates.

1. The total energy of an electron in ground state of hydrogen atom is -13.6eV. What is the significance of negative sign?
2. The radius of innermost electron orbit of hydrogen atom is 5.3 × 10¹¹m. What are the radii of n = 2 and n = 3 orbits?

Answer:
1. Negative sign implies that the electrons are strongly bounded to the nucleus.

2. r_n = n²a₀ = 5.3 × 10^-11m
r₁ = a₀ = 5.3 × 10^-11m
r₂ = 4a₀ = 21.2 × 10^-11m
r₃ = 9a₀ = 47.7 × 10^-11m.