Atomic Structure - NEET/JEE /KEAM
Each correct answer have 4 Marks Each wrong answers have -1 Mark Time - 15 Minutes ( 15 Questions )
Questions and Answers
- 1.
If the energy of a proton is given as 3.03 × 10-19 J then, the wave length of the photon is:
- A.
6.56nm
- B.
65.6 nm
- C.
656 nm
- D.
0.656 nm
Correct Answer
C. 656 nmExplanation
The wavelength of a photon can be determined using the equation E = hc/λ, where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the photon. Rearranging the equation to solve for λ, we get λ = hc/E. Plugging in the given values of E = 3.03 × 10-19 J, h = 6.626 × 10-34 J·s, and c = 3 × 108 m/s, we can calculate the wavelength to be 656 nm.Rate this question:
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- 2.
In the photo-electron emission, the energy of the emitted electron is
- A.
Greater than the incident photon
- B.
Same as than of the incident photon
- C.
Smaller than incident photon
- D.
Proportional to the intensity of incident photon
Correct Answer
C. Smaller than incident pHotonExplanation
In photo-electron emission, the energy of the emitted electron is smaller than that of the incident photon. This is because when a photon interacts with an electron, only a portion of its energy is transferred to the electron. The remaining energy is lost in the form of heat or other processes. Therefore, the energy of the emitted electron is less than the energy of the incident photon.Rate this question:
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- 3.
The electron was shown experimentally to have wave properties by
- A.
De Broglie
- B.
Davisson and Germer
- C.
N. Bohr
- D.
Schrodinger
Correct Answer
B. Davisson and GermerExplanation
Davisson and Germer showed experimentally that electrons have wave properties through their famous electron diffraction experiment in 1927. They directed a beam of electrons at a nickel crystal and observed a diffraction pattern, similar to the diffraction of light waves passing through a narrow slit. This confirmed the wave-particle duality of electrons, supporting the concept proposed by Louis de Broglie that particles like electrons can exhibit wave-like behavior. This experiment played a crucial role in the development of quantum mechanics and our understanding of the nature of matter.Rate this question:
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- 4.
Which of the following is never true for cathode rays?
- A.
They possess kinetic energy
- B.
They are electromagnetic waves
- C.
They produce heat
- D.
They produce mechanical pressure
Correct Answer
B. They are electromagnetic wavesExplanation
Cathode rays are not electromagnetic waves because they are actually streams of electrons. They possess kinetic energy because they are moving particles. They can produce heat and mechanical pressure when they collide with other objects, but they themselves are not electromagnetic waves.Rate this question:
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- 5.
According to the Bohr theory, which of the following transmissions in the hydrogen atom will give rise to the least energetic photon?
- A.
N= 6 to n=1
- B.
N=5 to n= 4
- C.
N= 6 to n= 5
- D.
N= 5 to n=3
Correct Answer
C. N= 6 to n= 5Explanation
According to the Bohr theory, the energy of a photon emitted or absorbed during a transition in the hydrogen atom is given by the equation E = -13.6 eV * (1/n^2). The transition with the least energetic photon will have the smallest change in energy. In this case, the transition from n=6 to n=5 has a smaller change in energy compared to the other options, resulting in the least energetic photon.Rate this question:
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- 6.
In hydrogen atom, energy of first excited state is -3.4 eV. Find out KE of the same orbit of Hydrogen atom
- A.
+3.4 eV
- B.
+6.8 eV
- C.
-13.6 eV
- D.
+13.6 eV
Correct Answer
A. +3.4 eVExplanation
The energy of an electron in an orbit of a hydrogen atom is given by the equation E = -13.6/n^2 eV, where n is the principal quantum number. The first excited state corresponds to n = 2. Plugging this value into the equation, we get E = -13.6/2^2 = -3.4 eV. Since the question asks for the kinetic energy (KE) of the same orbit, the KE would also be -3.4 eV.Rate this question:
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- 7.
According to Bohr's theory the energy required of for an electron in the Li​​​​​2+ ion to be emitted from n= 2 state is (given that the ground state ionization energy of hydrogen atom is 13.6 eV)
- A.
61.2 eV
- B.
13.6 eV
- C.
30.6 eav
- D.
10.2 eV
Correct Answer
C. 30.6 eavExplanation
.Rate this question:
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- 8.
The Bohr orbit radius for the hydrogen atom(n=1) is approximately 0.530 A. The radius for the first excited state (n=2) orbit is (in A)
- A.
0.13
- B.
1.06
- C.
4.77
- D.
2.12
Correct Answer
D. 2.12Explanation
The radius for the first excited state (n=2) orbit of the hydrogen atom can be calculated using the formula for the Bohr radius, which is given by r = 0.529 * n^2/Z, where r is the radius, n is the principal quantum number, and Z is the atomic number. In this case, n=2 and Z=1 for hydrogen. Plugging in these values into the formula, we get r = 0.529 * 2^2/1 = 0.529 * 4 = 2.12 A. Therefore, the correct answer is 2.12.Rate this question:
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- 9.
The radius of hydrogen atom in the ground state is 0.53 A. The radius of Li​​​​​2+ ion (atomic number= 3) in a similar state is,
- A.
0.17 A
- B.
0.265 A
- C.
0.53 A
- D.
1.06 A
Correct Answer
A. 0.17 AExplanation
In the ground state, the radius of an atom is determined by the size of its electron cloud. As the atomic number increases, the number of protons in the nucleus also increases, leading to a stronger attraction between the protons and the electrons. This stronger attraction causes the electron cloud to be pulled closer to the nucleus, resulting in a smaller radius. Since Li2+ has an atomic number of 3, it has one less electron than hydrogen. Therefore, the radius of Li2+ in a similar state would be smaller than that of hydrogen. Among the given options, 0.17 A is the smallest radius, indicating that it is the correct answer.Rate this question:
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- 10.
If ionization potential for hydrogen atom is 13.6 eV, then ionization potential for He+ will be,
- A.
54.4 eV
- B.
6.8 eV
- C.
13.6 eV
- D.
24.5 eV
Correct Answer
A. 54.4 eVExplanation
The ionization potential for an atom is the energy required to remove an electron from the atom. In the case of hydrogen, the ionization potential is 13.6 eV. When hydrogen loses an electron, it becomes a hydrogen ion (H+). However, in the given question, we are asked about the ionization potential for He+ (helium ion). He+ is a helium atom that has lost one electron. Since helium has two electrons, removing one electron requires more energy than removing one electron from hydrogen. Therefore, the ionization potential for He+ is higher than that of hydrogen. Among the given options, the only value that is higher than 13.6 eV is 54.4 eV, so that is the correct answer.Rate this question:
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- 11.
If r is the radius of the first orbit, the radius of n​​​th orbit of H- atom is given by
- A.
Rn​​​​​2
- B.
Rn
- C.
R/n
- D.
R​​​​​​2​​​n2
Correct Answer
A. Rn​​​​​2Explanation
The given answer, rn^2, is the correct formula for calculating the radius of the nth orbit of a hydrogen atom. This formula is derived from the Bohr model of the atom, which states that the radius of each orbit is proportional to the square of the principal quantum number, n. Therefore, multiplying the radius of the first orbit, r, by n^2 gives us the radius of the nth orbit.Rate this question:
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- 12.
The spectrum of He is expected to be similar to that
- A.
H
- B.
Li​​​+
- C.
Na
- D.
He+
Correct Answer
B. Li​​​+Explanation
The spectrum of He is expected to be similar to that of Li+ because both He and Li+ have similar electron configurations. He has two electrons in its 1s orbital, while Li+ has only one electron in its 1s orbital. This similarity in electron configuration results in similar energy levels and transitions, leading to similar spectra.Rate this question:
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- 13.
Two electrons occupying the same orbital are distinguished by
- A.
Principal quantum number
- B.
Magnetic quantum number
- C.
Azimuthal quantum number
- D.
Spin quantum number
Correct Answer
D. Spin quantum numberExplanation
Two electrons occupying the same orbital are distinguished by their spin quantum number. The spin quantum number describes the intrinsic angular momentum of an electron and can have two possible values: +1/2 or -1/2. This means that two electrons in the same orbital must have opposite spin values, allowing them to occupy the same space without violating the Pauli exclusion principle. The other quantum numbers (principal, magnetic, and azimuthal) are used to describe the energy level, orientation, and shape of the orbital, but they do not differentiate between electrons in the same orbital.Rate this question:
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- 14.
The correct set of four quantum numbers for the valence electron of rubidium atom (Z= 37) is,
- A.
5,1,+1/2
- B.
6,0,0 +1/2
- C.
5,0,0 +1/2
- D.
5,1, 0 +1/2
Correct Answer
C. 5,0,0 +1/2Explanation
The set of quantum numbers for an electron in an atom describes its energy level, orbital shape, orientation, and spin. The first quantum number, n, represents the energy level or shell of the electron. The second quantum number, l, represents the orbital shape or subshell of the electron. The third quantum number, ml, represents the orientation or magnetic quantum number of the electron within the subshell. The fourth quantum number, ms, represents the spin of the electron. In the given options, the correct set of quantum numbers for the valence electron of rubidium (Z=37) is 5,0,0,+1/2. This indicates that the valence electron is in the fifth energy level, s orbital (l=0), with no specific orientation (ml=0), and a spin of +1/2.Rate this question:
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- 15.
The total number of atomic orbitals in fourth energy level of an atom is:
- A.
8
- B.
16
- C.
32
- D.
4
Correct Answer
B. 16Explanation
The fourth energy level of an atom contains 4 sublevels: s, p, d, and f. The s sublevel has 1 orbital, the p sublevel has 3 orbitals, the d sublevel has 5 orbitals, and the f sublevel has 7 orbitals. Adding up the number of orbitals in each sublevel, we get 1 + 3 + 5 + 7 = 16. Therefore, the total number of atomic orbitals in the fourth energy level of an atom is 16.Rate this question:
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