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All questions of Structure of Atom for NEET Exam

How many subshells and electrons are associated with n = 4?
  • a)
    32, 64
  • b)
    16, 32
  • c)
    4, 16
  • d)
    8, 16
Correct answer is option 'B'. Can you explain this answer?

Jyoti Sengupta answered
For nth orbital possible values of azimuthal quantum number (subshell), l are from 0 to (n-1). Total of 'n' values.
In n=4, l=0,1,2,3 thus there are 4 subshells i.e.s,p,d,f respectively.
Magnetic quantum number ml have values from -l to +l and total of 2l+1 values.
For n=4, possible values of l and ml are:
ml=0 for l=0; total ml values =1
ml=−1,0,1 for l=1; total ml values =3
ml=−2,−1,0,1,2 for l=2; total ml values =5
ml=−3,−2,−1,0,1,2,3 for l=3; total ml values =7
Total number of orbitals = total values of ml
for n=4,
∴1+3+5+7=16 orbitals
Each orbital can occupy maximum of two electron
Number of electrons =2×16=32
Therefore in n=4, number of subshells=4, orbitals=16 and number of electrons =32.
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Describe the orbital with following quantum numbers:
(i) n = 3, l = 2
(ii) n = 4 , l = 3
  • a)
    (i) 3p, (ii) 4f
  • b)
    (i) 3d, (ii) 4d
  • c)
    (i) 3f, (ii) 4f
  • d)
    (i) 3d, (ii) 4f
Correct answer is option 'D'. Can you explain this answer?

Mira Joshi answered
(i) n = 3, l = 2 ⇒ 3 d
(ii) n = 4 , l = 3 ⇒ 4f
Quantum Numbers Description

i) n = 3, l = 2

- For the quantum numbers n = 3 and l = 2, the orbital is in the 3d subshell.
- The 3d orbital has a complex shape with two angular nodes and can hold a maximum of 10 electrons.
- Electrons in the 3d orbital have higher energy compared to those in the s and p orbitals.

ii) n = 4, l = 3

- For the quantum numbers n = 4 and l = 3, the orbital is in the 4f subshell.
- The 4f orbital has a more complex and intricate shape compared to lower energy orbitals.
- The 4f orbital can hold a maximum of 14 electrons and is located further from the nucleus due to higher energy levels.

Mark the incorrect statement regarding the photoelectric effect.
  • a)
    There is no time lag between the striking of light beam and the ejection of electrons from the metal surface
  • b)
    The number of electrons ejected is inversely proportional to the intensity of light
  • c)
    Photoelectric effect is not observed below threshold frequency
  • d)
    The kinetic energy of the electrons increases with increase in frequency of light used
Correct answer is option 'B'. Can you explain this answer?

Himanshu Das answered
B) the number of electrons ejected is directly proportional to the intensity of light as a certain amount threshold frequency is required for the photoelectric effect,,,a light with more intesity will contain mor threshold frequency than a low intensity light so the number of electrons ejected will depend upon the intensity of light,,,

What will be the energy of one photon of radiation whose frequency is 5 × 1014 Hz?
  • a)
    199.51J
  • b)
    3.3 x 10-19J
  • c)
    6.626 x 10-34 J
  • d)
    2.31 x 105 J
Correct answer is option 'B'. Can you explain this answer?

Suresh Iyer answered
The energy of one photon is given by the expression, E = hν
where, h = 6.626 × 10−34 Js and ν= 5 × 1014 s−1
E = 6.626 × 10−34 × 5 × 1014 = 3.313 × 10−19J

How many electrons in an atom have the following quantum numbers?
n = 4,  s = -1/2
  • a)
    32
  • b)
    18
  • c)
    8
  • d)
    16
Correct answer is option 'D'. Can you explain this answer?

Ananya Das answered
n = 4
l = 0, 1 , 2 , 3
ml = 0, (-1, 0, +1), (-2, -1, 0, +1, +2), (-3, -2, -1, 0, +1, +2, +3)

Total number of electrons with ms = - 1/2 will be 16. 

In how many elements the last electron will have the following set of quantum numbers, n = 3 and l = 1?
  • a)
    2
  • b)
    8
  • c)
    6
  • d)
    10
Correct answer is option 'C'. Can you explain this answer?

Ashwin Saini answered
Explanation:

The set of quantum numbers given are n=3 and l=1.

The principal quantum number (n) determines the energy level of the electron, while the azimuthal quantum number (l) determines the subshell in which the electron resides.

For l=1, the subshell is the p subshell. In the third energy level (n=3), there are three subshells: s, p, and d.

The maximum number of electrons that can occupy a p subshell is 6, according to the Pauli Exclusion Principle and Hund's Rule.

Therefore, the number of elements that can have the quantum numbers n=3 and l=1 (p subshell) is 6.

Option C is correct.

A body of mass 10 g is moving with a velocity of 100 m s-1. The wavelength associated with it is
  • a)
    6.626 x 10-7 m
  • b)
    6.626 x 10-34 m
  • c)
    6.626 x 10-4 m
  • d)
    6.626 x 10-35 m
Correct answer is option 'B'. Can you explain this answer?

Milan Nambiar answered
Explanation:
The given problem involves calculating the de Broglie wavelength associated with a body of mass and velocity.

De Broglie Wavelength Formula:
The de Broglie wavelength (λ) of a particle is given by the formula:
λ = h / mv
Where:
λ = de Broglie wavelength
h = Planck's constant (6.626 x 10^-34 J s)
m = mass of the particle (10 g = 0.01 kg)
v = velocity of the particle (100 m/s)

Calculation:
Substitute the given values into the formula:
λ = (6.626 x 10^-34) / (0.01 x 100)
λ = 6.626 x 10^-34 / 1
λ = 6.626 x 10^-34 m
Therefore, the de Broglie wavelength associated with the body of mass 10 g moving at 100 m/s is 6.626 x 10^-34 m.
So, the correct answer is option B: 6.626 x 10^-34 m.

What is the lowest value of n that allows g orbital to exist?
  • a)
    6
  • b)
    7
  • c)
    4
  • d)
    5
Correct answer is option 'D'. Can you explain this answer?

Saumya Sarkar answered
The lowest value of n that allows the g orbital to exist is 5.

Explanation:
The g orbital is one of the five d orbitals. It has a complex shape and can hold a maximum of 10 electrons. The d orbitals exist in the energy levels beyond the s and p orbitals.

Energy Levels and Subshells:
- The energy levels in an atom are represented by the principal quantum number (n).
- The value of n determines the size and energy of the orbital.
- The subshells, which are further divided into orbitals, are represented by the azimuthal quantum number (l).
- The value of l determines the shape of the orbital.

Order of Filling Orbitals:
The order of filling orbitals is determined by the Aufbau principle, which states that electrons fill the lowest energy level orbitals first before moving to higher energy level orbitals.

Order of Filling Orbitals:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p...

Explanation:
1. The s orbitals (spherical) are filled first, starting with the 1s orbital, followed by 2s, 3s, 4s, and so on.
2. The p orbitals (dumbbell-shaped) are filled after the s orbitals, starting with the 2p orbital, followed by 3p, 4p, and so on.
3. The d orbitals (complex shapes) are filled after the p orbitals, starting with the 3d orbital, followed by 4d, 5d, and so on.
4. The f orbitals (even more complex shapes) are filled after the d orbitals, starting with the 4f orbital, followed by 5f, and so on.

The g Orbital:
- The g orbital is the fifth orbital in the d subshell.
- It is filled after the 4s, 3d, and 4p orbitals.
- Therefore, the lowest value of n that allows the g orbital to exist is 5.

Answer:
The correct answer is option 'D' (5).

The probability of finding out an electron at a point within an atom is proportional to the
  • a)
    square of the orbital wave function i.e., ψ2
  • b)
    orbital wave function i.e., ψ
  • c)
    Hamiltonian operator i.e., H
  • d)
    principal quantum number i.e., n
Correct answer is option 'A'. Can you explain this answer?

Preeti Iyer answered
ψ2 is known as probability density and is always positive. From the value of ψ2 at different points with in an atom it is possible to predict the region around the nucleus where electron will most probably be found.

What will be the uncertainty in velocity of a bullet with a mass of 10 g whose position is known with ± 0.01 mm?
  • a)
    5.275 x 10-33 m s-1
  • b)
    5.275 x 10-25 m s-1
  • c)
    5.275 x 10-5 m s-1
  • d)
    5.275 x 10-28 m s-1
Correct answer is option 'D'. Can you explain this answer?

Arshiya Bajaj answered

Calculation of Uncertainty in Velocity of a Bullet

- Given:
- Mass of the bullet, m = 10 g = 0.01 kg
- Uncertainty in position, Δx = 0.01 mm = 0.01 x 10^-3 m = 1 x 10^-5 m
- Planck's constant, h = 6.626 x 10^-34 Js

- The uncertainty in velocity (Δv) of the bullet can be calculated using the Heisenberg Uncertainty Principle:
Δv * Δx ≥ h / (4π)

- Substituting the given values into the equation:
Δv * 1 x 10^-5 m ≥ 6.626 x 10^-34 Js / (4π)
Δv ≥ 6.626 x 10^-34 Js / (4π * 1 x 10^-5 m)
Δv ≥ 5.275 x 10^-28 m/s

Therefore, the uncertainty in velocity of the bullet is 5.275 x 10^-28 m/s, which is option 'D'. This value represents the minimum uncertainty in the velocity of the bullet due to the uncertainty in its position, as described by the Heisenberg Uncertainty Principle.

What does the negative electronic energy (negative sign for all values of energy) for hydrogen atom means?
  • a)
    The energy of an electron in the atom is lower than the energy of a free electron at rest which is taken as zero
  • b)
    When the electron is free from the influence of nucleus it has a negative value which becomes more negative
  • c)
    When the electron is attracted by the nucleus the energy is absorbed which means a negative value
  • d)
    Energy is released by hydrogen atom in ground state
Correct answer is option 'A'. Can you explain this answer?

Ujwal Sen answered
Understanding Negative Electronic Energy in Hydrogen Atom
The concept of negative electronic energy in the hydrogen atom is fundamental to quantum mechanics and atomic theory. Let's explore what this means:
Energy Reference Point
- In atomic physics, the energy of a free electron at rest is considered as the reference point and is assigned a value of zero.
- When an electron is bound to a nucleus (like in a hydrogen atom), its energy is negative, indicating that it is in a lower energy state compared to when it is free.
Lower Energy State
- The negative sign of the energy indicates that the electron is in a more stable configuration when it is bound to the nucleus.
- This means that the electron has less energy than when it is not influenced by the nucleus.
Comparison with Free Electron
- A negative energy value suggests that energy must be added to the system to free the electron from the attractive force of the nucleus.
- Thus, when moving from a bound state to a free state, the energy of the electron transitions from negative to zero, indicating energy absorption.
Conclusion
- Therefore, the correct interpretation is that the energy of an electron in the hydrogen atom is lower than that of a free electron at rest, which is taken as zero. This aligns with option 'A'.
- Understanding this helps in realising the stability of electrons in atoms and their behavior in different energy states.

What will be the orbital angular momentum of an electron in 2s-orbital?
  • a)
    Zero
  • b)
    One
  • c)
    Two
  • d)
    Three
Correct answer is option 'A'. Can you explain this answer?

Ameya Bose answered
Introduction:
The orbital angular momentum of an electron is a property that arises from its motion around the nucleus in an atom. It quantifies the amount of rotational motion an electron has in its orbit. The orbital angular momentum is determined by the principal quantum number (n) and the azimuthal quantum number (l) of the orbital in which the electron resides.

Explanation:
- The principal quantum number, n, describes the energy level of an electron and determines the size of the orbital.
- The azimuthal quantum number, l, describes the shape of the orbital and can take on values from 0 to n-1.
- For an s-orbital, the azimuthal quantum number, l, is equal to 0. Therefore, the orbital angular momentum (L) of an electron in an s-orbital is given by the equation L = √(l(l+1))ħ, where ħ is the reduced Planck's constant.
- Plugging in the value l = 0 into the equation, we get L = √(0(0+1))ħ = √(0)ħ = 0.
- Hence, the orbital angular momentum of an electron in a 2s-orbital is zero.

Conclusion:
The orbital angular momentum of an electron in a 2s-orbital is zero. This is because the azimuthal quantum number, l, for an s-orbital is zero, resulting in a zero value for the orbital angular momentum.

Which of the following configurations represents a noble gas?
  • a)
    1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2
  • b)
    1s2 2s2 2p6 3s2 3p6 3d10 4s2 4f14 5s2
  • c)
    1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6
  • d)
    1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p3
Correct answer is option 'C'. Can you explain this answer?

Janhavi Patel answered

Explanation:

Noble Gas Configuration:

- A noble gas configuration is when an atom has achieved a stable electron configuration similar to that of a noble gas.
- Noble gases have a completely filled outermost energy level, which makes them very stable and unreactive.

Identifying the Noble Gas Configuration:

- The electron configuration given in option 'C' is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2.
- This configuration represents the noble gas Krypton (Kr), which has the electron configuration of [Ar] 4s2 3d10 4p6.
- By comparing the given configuration with the electron configuration of Krypton, we can see that it matches, indicating a noble gas configuration.

Conclusion:

- Option 'C' (1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2) represents a noble gas configuration similar to that of Krypton, making it the correct answer.

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