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# Test: Atoms (CBSE Level With Solutions)

## 30 Questions MCQ Test Physics Class 12 | Test: Atoms (CBSE Level With Solutions)

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This mock test of Test: Atoms (CBSE Level With Solutions) for Class 12 helps you for every Class 12 entrance exam. This contains 30 Multiple Choice Questions for Class 12 Test: Atoms (CBSE Level With Solutions) (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Atoms (CBSE Level With Solutions) quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test: Atoms (CBSE Level With Solutions) exercise for a better result in the exam. You can find other Test: Atoms (CBSE Level With Solutions) extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.
QUESTION: 1

### According to ‘plum pudding model’ atoms on the whole are electrically neutral because

Solution:

The plum pudding model is one of several scientific models of the atom. According to J.J. Thomson atomic models the positive particles in the atom form something like the "batter" in a plum pudding, while the negative electrons are scattered through this "batter".

QUESTION: 2

### Emission line spectra of different elements

Solution:

A spectrum is an assembly of energy levels in the form of radiations emitted by an atom in its excited state. Every atom gives discontinuous line spectra. Each line in the spectra corresponds to a specific wavelength and it is unique to a given element so no two elements give the same pattern of lines in their spectra.

QUESTION: 3

### What is the shortest wavelength present in the Paschen series of spectral lines?

Solution:

Using formula for Paschen series,
1/λ=R[(1/32)-1/n22],n2=4,5,6….
For shortest wavelength, n2=∞
∴1/λ=R[(1/32)-1/∞2],n2=R/9 or λ=9/R ≈ 820.4 nm

QUESTION: 4

Average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model is

Solution:

The average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

QUESTION: 5

The average lifetime of the first excited level of a hydrogen atom is 1.0 ×10−8 s. In the Bohr model, how many orbits does an electron in the n = 2 level complete before returning to the ground level?

Solution:  QUESTION: 6

Each element is associated with

Solution:
QUESTION: 7

Which of these statements about Bohr model is correct?

Solution:

Bohr model of the hydrogen atom attempts to plug in certain gaps as suggested by Rutherford’s model by including ideas from the newly developing Quantum hypothesis. According to Rutherford’s model, an atom has a central nucleus and electron/s revolve around it like the sun-planet system.
However, the fundamental difference between the two is that, while the planetary system is held in place by the gravitational force, the nucleus-electron system interacts by Coulomb’s Law of Force. This is because the nucleus and electrons are charged particles. Also, an object moving in a circle undergoes constant acceleration due to the centripetal force.
Further, electromagnetic theory teaches us that an accelerating charged particle emits radiation in the form of electromagnetic waves. Therefore, the energy of such an electron should constantly decrease and the electron should collapse into the nucleus. This would make the atom unstable.
The classical electromagnetic theory also states that the frequency of the electromagnetic waves emitted by an accelerating electron is equal to the frequency of revolution. This would mean that, as the electron spirals inwards, it would emit electromagnetic waves of changing frequencies. In other words, it would emit a continuous spectrum. However, actual observation tells us that the electron emits a line spectrum.

QUESTION: 8

Which of these statements about Bohr model hypothesis is correct?

Solution:

Bohr never assumed stable electron orbits with the electronic angular momentum quantized as l=mvr=(nh/2π)​ Quantization of angular momentum means that the radius of the orbit and the energy will be quantized as well. Bohr assumed that the discrete lines seen in the spectrum of the hydrogen atom were due to transitions of an electron from one allowed orbit/energy to another.

QUESTION: 9

Which of these statements about Bohr model applied to hydrogen atom correct?

Solution:

A hydrogen atom has magnetic properties because the motion of the electron acts as a current loop. The energy levels of a hydrogen atom associated with orbital angular momentum are split by an external magnetic field because the orbital angular magnetic moment interacts with the field.

QUESTION: 10

Probability of backward scattering (i.e., scattering of α -particles at angles greater than 90∘) predicted by Thomson’s model is

Solution:

In Rutherford's model we have a large massive core called the nucleus whereas, in Thomson's model we do not have. Thus the probability of backward scattering by Thomson's model is much less than that predicted by Rutherford model.

QUESTION: 11

In Geiger-Marsden experiment very small deflection of the beam was expected because

Solution:

In the Geiger-Marsden experiment very small deflection of the beam was expected because positive charge and the negative electrons are distributed through the whole atom reducing electric field inside the atom.

QUESTION: 12

Fluorescence is

Solution:

Fluorescence is the emission of light by a substance that has absorbed light or other electromagnetic radiation. It is a form of luminescence. In most cases, the emitted light has a longer wavelength, and therefore lower energy, than the absorbed radiation. The most striking example of fluorescence occurs when the absorbed radiation is in the ultraviolet region of the spectrum, and thus invisible to the human eye, while the emitted light is in the visible region, which gives the fluorescent substance a distinct color that can be seen only when exposed to UV light. Fluorescent materials cease to glow nearly immediately when the radiation source stops, unlike phosphorescent materials, which continue to emit light for some time after.

QUESTION: 13

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Solution:

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67×10−27) is less than the mass of incident α−particles (6.64×10−27). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α−particle scattering experiment.

QUESTION: 14

It is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

Solution:
QUESTION: 15

An electron collides with a hydrogen atom in its ground state and excites it to a state of n = 3. How much energy was given to the hydrogen atom in this inelastic collision?

Solution:

Energy at ground state E1​=−13.6 eV
Energy at n=3: E3​=−13.6/9​=1.5 eV
To excite it to n=3 energy given to electron is E3​−E1​=12.1 eV

QUESTION: 16

Which of these statements correctly describe the atomic model according to classical electromagnetic theory ?

Solution:

In classical electromagnetic theory, atoms and molecules are considered to contain electrical charges (i.e. electrons, ions) which are regarded as oscillating about positions of equilibrium, each with its appropriate natural frequency, v0 . When placed in a radiation field of frequency v , each oscillator in the atom or molecule is set into forced vibration with the same frequency as that of the radiation. The amplitude of the forced vibration is small, but as v approaches v0 , the amplitude of the forced vibration increases rapidly. To account for the absorption of energy from the radiation field, it is necessary to assume that the oscillator in the atom or molecule must overcome some frictional force proportional to its velocity during its forced motion. For small amplitudes of forced oscillation, the frictional force, and therefore the absorption of energy, is negligible. Near resonance , the amplitude of oscillation becomes large, with a correspondingly large absorption of energy to overcome the frictional force. Therefore, the radiation of frequencies near the natural frequency of the oscillator corresponds to an absorption band.

QUESTION: 17

Reason why there are many lines in an atomic spectrum is because

Solution:

Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states.

QUESTION: 18

In the ground state of which model electrons are in stable equilibrium with zero net force?

Solution:

In Thomson's model, the atom is composed of electrons surrounded by a soup of positive charge to balance the electrons' negative charges, like negatively charged “plums” surrounded by positively charged “pudding”. The 1904 Thomson model was disproved by Hans Geiger and Ernest Marsden's 1909 gold foil experiment.

QUESTION: 19

A triply ionized beryllium ion Be3+, (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. For the hydrogen atom, the wavelength of the photon emitted in the n =2 to n=1 to transition is 122 nm. What is the wavelength of the photon emitted when a Be3+ ion undergoes this transition?

Solution:

1/ λ =Z2. (both have same transition so same value of n)
λ /122=1/16
λ =122/16
=7.62nm

QUESTION: 20

In Geiger-Marsden experiment, at the point of closest approach

Solution:

In the Geiger-Marsden experiment, at the point of closest approach the kinetic energy is zero and the electrical potential equals the initial kinetic energy supplied.

QUESTION: 21

Fluorescent lamps are more efficient than incandescent lamps in converting electrical energy to visible light because

Solution:

The phosphor fluoresces to produce light. A fluorescent bulb produces less heat, so it is much more efficient. This makes fluorescent bulbs four to six times more efficient than incandescent bulbs. That's why you can buy a 15-watt fluorescent bulb that produces the same amount of light as a 60-watt incandescent bulb.

QUESTION: 22

In which of the models An atom has a nearly continuous mass distribution?

Solution:

An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.

QUESTION: 23

Find the longest wavelength present in the Balmer series of hydrogen, corresponding to the H- line.

Solution:

For single electron species,
1/λ​=RZ2[(1/n12​​)−(1/n22​​])
where, R = Rydberg constant =  1.097×107×10−9nm−1=1.097×10−2nm−1
For Hydrogen atom, Z = Atomic number = 1
For Balmer series, n1​ = 2 and for longest wavelength in Balmer series, minimum energy transition is to be considered because wavelength is inversely proportional to Energy.
So, n2​ = 3
∴1/λ​=1.097×10−2nm−1×12[(1/22)​−(1/32)]
⇒λ=656nm

QUESTION: 24

According to Bohr model radiation takes place when

Solution:

In 1913 Bohr proposed his quantized shell model of the atom to explain how electrons can have stable orbits around the nucleus. The motion of the electrons in the Rutherford model was unstable because, according to classical mechanics and electromagnetic theory, any charged particle moving on a curved path emits electromagnetic radiation; thus, the electrons would lose energy and spiral into the nucleus. To remedy the stability problem, Bohr modified the Rutherford model by requiring that the electrons move in orbits of fixed size and energy. The energy of an electron depends on the size of the orbit and is lower for smaller orbits. Radiation can occur only when the electron jumps from one orbit to another. The atom will be completely stable in the state with the smallest orbit, since there is no orbit of lower energy into which the electron can jump.
hν = Ef−Ei

QUESTION: 25

The model that best explains the results of Geiger-Marsden experiment is

Solution:

When Rutherford saw the results of the experiment by Geiger and Marsden, he said:
“It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”
Rutherford used the results of this experiment to develop a new model for the atom. This model proposed a central nucleus with a positive charge. It was this positively charged nucleus that was responsible for the strong backward deflection of the positively charged alpha particles.
The model also proposed that negatively charged electrons surrounded this nucleus. However, as most of the alpha particles passed through the gold foil with no deflection at all, Rutherford realised that most of the atom was empty space. So, his model placed the electrons at some distance from the nucleus.

QUESTION: 26

To produce an emission spectrum of hydrogen

Solution:
QUESTION: 27

According to Bohr model allowed values of angular momentum are

Solution:

The angular momentum (L) of an electron in a Bohr orbit is given as L= nh /2π ​. It is an integral multiple of h/2π​.

QUESTION: 28

Which of these statements about Bohr model hypothesis is correct?

Solution:
QUESTION: 29

In a Geiger -Marsden experiment, what is the distance of closest approach d to the nucleus of a 7.7 MeV α−particle before it comes momentarily to rest and reverses its direction?

Solution:

The distance of closest approach is given as
r0= (1/4πε0)(2Ze2/E)
Here,
Z= 75
e = 1.6x10-19 C
E = 5 MeV = 5 X 1.6 X 10-13 J
1/4πε0 = 9x109 Nm2C-1
so,
r0 = [9x109 x 2x75x(1.6x10-19)2] / [5 X 1.6 X 10-13] m
r0=30x10−15 m
r0=30

QUESTION: 30

Absorption line spectrum is obtained

Solution:

If light from a continuous spectrum passes through a cool, transparent gas we observe dark lines appear in the spectrum. The lines occur where atoms of the gas have absorbed specific wavelengths of light. Hence we call this type of spectrum an absorption spectrum.