Atomic Spectra MCQ Quiz in தமிழ் - Objective Question with Answer for Atomic Spectra - இலவச PDF ஐப் பதிவிறக்கவும்

²³⁸U → ₉₀Th²³⁴ + ₂He⁴

An alpha particle has a mass number and atomic number equal to He.

Concept:

Nuclear Magnetic Resonance Spectroscopy (NMR) test

• Nuclear magnetic resonance (NMR) spectroscopy is the study of molecules by recording the interaction of radiofrequency (Rf) electromagnetic radiations with the nuclei of molecules placed in a strong magnetic field.
• It is a research technique that exploits the magnetic properties of certain atomic nuclei.
• The NMR spectroscopy determines the physical and chemical properties of atoms or molecules.
• It is a test that can ascertain the composition of a product at the molecular level.

Explanation:

• There are two peaks because there are two different environments for the hydrogens - in the CH₃ group and attached to the oxygen in the CHO group.
• They are in different places in the spectrum because they need slightly different external magnetic fields to bring them in to resonance at a particular radio frequency.

​Additional Information

Why NMR?

• Honey has a very distinct spectrum when run through a magnetic resonance system.
• We can compare those spectra to other spectra of similar honey to determine whether it has been adulterated with inverted sugars.
• Additionally, we can look at quality control products, such as whether it has been fermented.

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Concept:

Spectrum is the impression produced on a photographic film when the radiation (s) of a particular wavelength (s) is (are) analysed through a prism or diffraction grating.

Types of spectrum           

(1) Emission spectrum:

The spectrum produced by the emitted radiation is known as an emission spectrum. This spectrum corresponds to the radiation emitted (energy evolved) when an excited electron returns back to the ground state.           

(i) Continuous spectrum: When sunlight is passed through a prism, it gets dispersed into continuous bands of different colours. If the light of an incandescent object resolved through prism or spectroscope, it also gives a continuous spectrum of colours.           

(ii) Line spectrum: If the radiation obtained by the excitation of a substance is analysed with help of a spectroscope a series of thin bright lines of specific colours are obtained. There is dark space in between two consecutive lines. This type of spectrum is called a line spectrum or atomic spectrum.  

(2) Absorption spectrum:

The spectrum produced by the absorbed radiations is called absorption spectrum.

The hydrogen spectrum is an example of line emission spectrum or atomic emission spectrum.

The Energy of the various spectral lines (n₂ → n₁) is given by 

$E=hcR{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Calculation:

Given spectral line emitted for the transition n = 6 to n = 3 of Li²⁺ is equal to transition in H atom;

Let the transition in H atom be from n₂ → n₁;

For Li²⁺, Energy of the transition is given by 

$E=hcR\times 3^2\left[\frac{1}{3^2}-\frac{1}{6^2}\right]=hcR\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$

Now for the Transition in H atom,

$E=hcR\times 1^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]=hcR\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Equating both the energies,

⇒ n₁ = 1, n₂ = 2

Calculation:

E₁ = hc / λ = 2.178 × 10^-18 × z² × [1 / 1²]

E₂ = hc / λ = 2.178 × 10^-18 × z² × [1 / 2²]

So, E = E₂ - E₁

E = hc / λ = 2.178 × 10^-18 × z² × [1 / 1² - 1 / 2²]

⇒ λ = 1.214 × 10^-7 m

The valence electron is present in $5s^1$ subshell.

Hence, the set of quantum numbers are $n=5,l=0,m=0$ and $s=+{\displaystyle \frac{1}{2}}$.

Hence, the correct option is C

The total number of orbitals associated with the principal quantum number 5 is 25.

Hence, the correct option is $A$

Calculation:

Kinetic energy = eV = (1/2) m u²

u = √(2eV / m) ......(1)

Here, u is the velocity of the electron.

The de Broglie wavelength is λ = h / (m u)

h / λ = m u ......(2)

Substitute equation (1) in equation (2):

h / λ = m √(2eV / m) = √(2meV)

Hence, the correct option is C

Given :

Hole concentration $n_h=5\times {10}^{19}{m}^{-3}$

Hole mobility ${\mu }_h=0.01m^2{v}^{-1}{s}^{-1}$

Electron concentration $n_e=5\times {10}^{18}{m}^{-3}$

Electron mobility ${\mu }_e=2m^2{v}^{-1}{s}^{-1}$

Conductivity of semiconductor $\sigma =e(n_e{\mu }_e+n_h{\mu }_h)$

$\therefore \sigma =(1.6\times {10}^{-19})[(5\times {10}^{18})(2)+(5\times {10}^{19})(0.01)]$

$⟹\sigma =1.68(\mathrm{\Omega }-m{)}^{-1}$

Calculation:

r = n² h² / (4 π² m e² k)

r = (2² (6.6262 × 10⁻³⁴ Js)²) / (4 × (3.1416)² × 9.1091 × 10⁻³¹ kg × (1.60210 × 10⁻¹⁹ C)² × 9 × 10⁹)

r = 2.12 × 10⁻¹⁰ m

r = 2.12 Å

Hence, the option (C) is the correct answer.

Calculation:

The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is 2π × 0.529 Å.

First Bohr orbit of 'H' atom has radius r = 0.529 Å

Also, the angular momentum is quantised.

mvr = h / 2π

2πr = h / mv = λ

λ = 2π × 0.529 Å

So, the correct option is B.