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NEET • PHYSICS • ATOMS

Atoms practice · Mixed difficulty

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Showing 31-60 of 130 questions

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31 Medium
31 of 130
Application +4 | -1
The ratio of ionization energy of Bohr’s hydrogen atom and Bohr’s hydrogen like lithium atom is
View answer and explanation
Correct option: C
1:9\mathrm{1 : 9}
Option C (1:9\mathrm{1 : 9}) is correct because ionization energy is proportional to Z2Z^2, and lithium has Z=3 while hydrogen has Z=1.
32 Easy
32 of 130
Remembering +4 | -1
The path of scattered α\alpha-particles is
View answer and explanation
Correct option: D
Hyperbolic
Option D (Hyperbolic) is correct because the path of scattered alpha particles in Rutherford scattering is a hyperbola due to the repulsive electrostatic force.
33 Medium
33 of 130
Understanding +4 | -1
In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is
View answer and explanation
Correct option: C
−1/2\mathrm{ - 1/2}
Option C (−1/2\mathrm{ - 1/2}) is correct because the kinetic energy is the negative half of the potential energy in a Bohr orbit.
34 Medium
34 of 130
Understanding +4 | -1
Assertion (A): In Rutherford experiment only few α−\alpha- particles get rebounded.

Reason (R): Most of the atomic space is empty.
View answer and explanation
Correct option: B
Both (A) and (R) are true and (R) is not the correct explanation of (A)
Option B (Both (A) and (R) are true and (R) is not the correct explanation of (A)) is correct because while both the assertion and the reason are true, the reason doesn't directly explain why only a few alpha particles rebound. The rebounding is due to direct collisions with the nucleus, which is a small target.
35 Medium
35 of 130
Understanding +4 | -1
Statement-I: The trajectory traced by an incident particle depends on the impact parameter of collisions.

Statement-II: The impact parameter is the perpendicular distance of the initial velocity vector of the incident particle from the centre of the target nucleus.
View answer and explanation
Correct option: A
Statement-I and Statement-II are true
Option A (Statement-I and Statement-II are true) is correct because the trajectory of an incident particle in scattering depends on the impact parameter, which is defined as the perpendicular distance of the initial velocity vector from the center of the target nucleus.
36 Medium
36 of 130
Understanding +4 | -1
Number of spectral lines in hydrogen atom is
View answer and explanation
Correct option: D
Infinite
Option D (Infinite) is correct because the number of spectral lines in a hydrogen atom is infinite, as electrons can transition between any two energy levels.
37 Medium
37 of 130
Application +4 | -1
Work required to pull an electron from helium atom is 40 eV. Then the work required to pull both the electrons from the helium atom is
View answer and explanation
Correct option: D
94.4 eV
Option D (94.4 eV) is correct because after removing the first electron (40 eV), the effective nuclear charge increases, making it harder to remove the second electron. The energy required to remove the second electron is calculated using the formula for ionization energy of a hydrogen-like atom.
38 Hard
38 of 130
Analyzing +4 | -1
The radiation emitted when an electron jumps from 4th orbit to 2nd orbit in hydrogen like atom is made to incident on a metal surface to produce photoelectrons. In this way the ejected electrons are made to enter into a uniform magnetic field of strength 5 mT\sqrt{5} \,\mathrm{mT} and perpendicular to the direction of motion of the electrons. In the field electrons move on a circular path of radius 4.55 mm4.55 \mathrm{~mm}. If the work function of the metal is 1.1 eV1.1 \,\mathrm{eV}, then
(a) Atomic number of the atom.
(b) Stopping potential of the photoelectrons. Find a+b.a+b.
View answer and explanation
Correct option: A
13
Option A (13) is correct because the atomic number and stopping potential can be calculated from the given information and their sum is 13.
39 Medium
39 of 130
Analyzing +4 | -1
Assertion (A): Electrons in the atom are held due to coulomb forces.
Reason (R): The atom is stable only because the centripetal force due to Coulomb's law is balanced by the centrifugal force.
View answer and explanation
Correct option: C
Assertion is true but reason is false
Option A (41) is correct because the atomic number can be determined from the given potential differences and the relationship between the KαK_\alpha wavelength and the minimum wavelength of continuous X-rays.
40 Easy
40 of 130
Application +4 | -1
According to Bohr model, the diameter of first orbit of hydrogen atom will be
View answer and explanation
Correct option: B
1A01{A^0}
Option B (1A01{A^0}) is correct because the diameter is twice the Bohr radius, which is approximately 0.529 Å.
41 Medium
41 of 130
Understanding +4 | -1
Solid targets of different elements are bombarded by highly energetic electron beams. The frequency (f)\left( f \right) of the characteristic X-rays emitted from different targets varies with atomic number ZZ as
View answer and explanation
Correct option: B
f∝Z2f \propto {Z^2}
Option B (f∝Z2f \propto {Z^2}) is correct because according to Moseley's law, the frequency of characteristic X-rays is proportional to the square of the atomic number (Z).
42 Easy
42 of 130
Application +4 | -1
The time period of electron in the ground state of hydrogen atom is three times the time period of the electron in the second excited state of a certain hydrogen like atom (atomic number ZZ). The value of ZZ is
View answer and explanation
Correct option: D
9
Option D (9) is correct because the time period is proportional to n3/Z2, and using the given information, we can solve for Z.
43 Medium
43 of 130
Application +4 | -1
When an α\mathrm{\alpha } - particle of mass ‘m’ moving with the velocity ‘v’ bombards on a heavy nucleus of charge ‘Ze’, its distance of closest approach from the nucleus depends on m as:
View answer and explanation
Correct option: A
1m\frac{1}{\mathrm{m}}
Option A (1m\frac{1}{\mathrm{m}}) is correct because the distance of closest approach is inversely proportional to the mass of the alpha particle.
44 Medium
44 of 130
Analyzing +4 | -1
Consider a hydrogen like atom whose energy in nth \mathrm{n}^{\text {th }} excited state is given by

En=−13.6n2Z2E_n=-\frac{13.6}{n^2} Z^2
when this excited atom makes a transition from an excited state to ground state. The most energetic photons have energy Emax⁡=52.224eV\mathrm{E}_{\max }=52.224 \mathrm{eV} and the least energetic photons have energy Emin⁡=1.224eV\mathrm{E}_{\min }=1.224 \mathrm{eV}. Find the atomic number of atom.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the atomic number can be determined from the given energy values using the formula for energy levels in a hydrogen-like atom.
45 Medium
45 of 130
Analyzing +4 | -1
Stopping potentials of 24,100,110,115kV24,100,110,115 \mathrm{kV} are measured for photoelectrons emitted from a certain element when it is irradiated with monochromatic X-rays. The element is used as a target in an X\mathrm{X}-ray tube. The energy of K\mathrm{K} line is -
View answer and explanation
Correct option: B
76KeV76 \mathrm{KeV}
Option B (76 KeV) is correct because the energy of the K line corresponds to the difference between the stopping potentials.
46 Hard
46 of 130
Application +4 | -1
An α\alpha-particle after passing through a potential difference of 2×106 V2 \times 10^6 \mathrm{~V} falls on a silver foil. The atomic number of silver is 47. Calculate (i) the kinetic energy of the α\alpha-particle at the time of falling on the foil (ii) the kinetic energy of the α\alpha-particle at a distance of 5×10−14 m5 \times 10^{-14} \mathrm{~m} from the nucleus and (iii) the shortest distance from the nucleus of silver to which the α\alpha-particle reaches.
View answer and explanation
Correct option: A
6.4×10−13 J; 2.1×10−13 J; 3.4×10−14 m6.4 \times 10^{-13} \mathrm{~J};\,2.1 \times 10^{-13} \mathrm{~J};\,3.4 \times 10^{-14} \mathrm{~m}
Option A (6.4×10−13 J; 2.1×10−13 J; 3.4×10−14 m6.4 \times 10^{-13} \mathrm{~J};\,2.1 \times 10^{-13} \mathrm{~J};\,3.4 \times 10^{-14} \mathrm{~m}) is correct because it accurately calculates the kinetic energies and shortest distance based on the potential difference and atomic number.
47 Easy
47 of 130
Understanding +4 | -1
In an atom for the electron to revolve around the nucleus, the necessary centripetal force is obtained from the following force exerted by the nucleus on the electron
View answer and explanation
Correct option: D
Electrostatics force
Option B (PEPE increases, TETE increases) is correct because as the electron moves to a higher orbit, its potential energy becomes less negative (increases) and its total energy also increases (becomes less negative).
48 Medium
48 of 130
Analyzing +4 | -1
In XX-ray tube when the accelerating voltage VV is halved, the difference between the wavelength of Kα{K_\alpha } line and minimum wavelength of continuous XX-ray spectrum
View answer and explanation
Correct option: D
Becomes less than two times
Option D (Becomes less than two times) is correct because halving the voltage affects the minimum wavelength and the difference between the wavelengths.
49 Medium
49 of 130
Application +4 | -1
The extreme wavelengths of Paschen series are
View answer and explanation
Correct option: B
0.818  μm0.818\,\,\mu m and 1.89  μm1.89\,\,\mu m
Option B (0.818μm0.818 \mu m and 1.89μm1.89 \mu m) is correct because these are the calculated extreme wavelengths for the Paschen series using the Rydberg formula.
50 Medium
50 of 130
Understanding +4 | -1
An electron with kinetic energy 5  eV5\,\,eV is incident on a HH-atom in its ground state. The collision
View answer and explanation
Correct option: A
Must be elastic
Option A (Must be elastic) is correct because the electron's kinetic energy is less than the energy required to excite the hydrogen atom to the next energy level.
51 Medium
51 of 130
Application +4 | -1
The first line of Balmer series has wavelength 6563  A06563\,\,{A^0}. What will be the wavelength of the first member of Lyman series
View answer and explanation
Correct option: A
1215.4  A01215.4\,\,{A^0}
Option A (1215.4  A01215.4\,\,{A^0}) is correct. Using the Rydberg formula, we can relate the wavelengths of the Balmer and Lyman series to find the wavelength of the first Lyman line.
52 Easy
52 of 130
Application +4 | -1
The hydrogen atoms in its ground state are excited by means of monochromatic radiation of wavelength 1023A∘1023 \mathop A\limits^ \circ. The number of lines which can be seen in the resulting spectrum is:
View answer and explanation
Correct option: A
3
Option A (3) is correct because the hydrogen atom can transition from n=3 to n=2, n=3 to n=1, and n=2 to n=1, resulting in 3 spectral lines.
53 Easy
53 of 130
Application +4 | -1
The potential difference across Coolidge tube is 20kv20 \mathrm{kv} and 10 mA10 \mathrm{~mA} current flows through the voltage supply. Only 0.5%0.5 \% of the energy carried by electrons striking the target is converted into X-ray. Then the power carried by X-ray beam in watts is -
View answer and explanation
Correct option: A
1
Option A (1) is correct because the power carried by the X-ray beam is 0.5% of the total power supplied to the tube, which is 20 kV * 10 mA = 200 W, and 0.5% of 200 W is 1 W.
54 Medium
54 of 130
Application +4 | -1
The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is
View answer and explanation
Correct option: D
90011\frac{{900}}{{11}}
Option D (90011\frac{{900}}{{11}}) is correct. The longest wavelength corresponds to the smallest energy difference, and the shortest wavelength corresponds to the largest energy difference across all series.
55 Medium
55 of 130
Remembering +4 | -1
For principal quantum number n=3n = 3, the possible values of orbital quantum number ′l′'l' are
View answer and explanation
Correct option: C
0,1,2\mathrm{0, 1, 2}
Option C (0, 1, 2) is correct because for a principal quantum number n=3, the orbital quantum number l can have values from 0 to n-1.
56 Medium
56 of 130
Remembering +4 | -1
The spectral series of the hydrogen atom that lies in the visible region of the electromagnetic spectrum
View answer and explanation
Correct option: B
Balmer
Option D (6) is correct because the number of emission transitions is calculated using the formula n(n-1)/2, where n is the highest energy level.
57 Medium
57 of 130
Understanding +4 | -1
The absorption transitions between the first and the fourth energy states of hydrogen atom are 3\mathrm{3}. The emission transitions between these states will be
View answer and explanation
Correct option: D
6\mathrm{6}
Option D (6) is correct because the number of emission transitions from energy level n to lower levels is given by n(n-1)/2. In this case, n=4, so 4(3)/2 = 6.
58 Medium
58 of 130
Understanding +4 | -1
Assertion: Rayleigh scattering can be considered as elastic collisions of photons with massive particles.
Reason: In Rayleigh scattering, the energy of incident and scattered is same
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because Rayleigh scattering involves elastic collisions where the energy (and therefore wavelength) of the incident and scattered photons are the same.
59 Easy
59 of 130
Remembering +4 | -1
The value of angular momentum for He+\mathrm{He}^{+} ion in the first Bohr orbit is:
View answer and explanation
Correct option: A
h2π\frac{h}{2 \pi}
Option A (h2π\frac{h}{2 \pi}) is correct because the angular momentum of an electron in the first Bohr orbit is always h2π\frac{h}{2 \pi}, regardless of the ion.
60 Medium
60 of 130
Application +4 | -1
The impact parameter at which the scattering angle is 90o, Z=79{90^\mathrm{o}},\,Z = 79 and initial energy 10 MeV is
View answer and explanation
Correct option: B
1.137×10−14m1.137 \times {10^{ - 14}}m
Option B (1.137×10−14m1.137 \times {10^{ - 14}}m) is correct because it is the impact parameter calculated using the formula for scattering angle of 90 degrees.
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