Q: From the $ \alpha $ -particle scattering experiment, Rutherford concluded that: This question has multiple correct options

the radius of the nucleus is less than $ 10^{-14} m $. Rutherford’s scattering showed that most of the atom is empty space and that the positive charge is concentrated in a tiny central nucleus. The existence of strong deflections at very small impact parameters implies the nucleus must have a radius smaller than about $10^{-14}\,\text{m}$.

Q: When a radioactive element emits successively one α-particle and two β-particles, the mass number of the daughter element:

is reduced by 4 units. Emission of an α-particle reduces the mass number by 4, while β-particle emissions do not affect the mass number. Thus, the mass number decreases by 4 units overall.

Q: If species ᵇₐX emits firstly a positron, then two α and two β and in last one α and finally converted to species ᶜₐY, so correct relation is:

c = a - 5, d = b - 8. The positron emission decreases the atomic number by 1, each α-particle emission decreases the atomic number by 2 and mass number by 4, and β-particle emission does not affect the mass number but increases the atomic number by 1. After applying these changes step-by-step, the final relation is c = a - 5 and d = b - 8.

Question 1

Protons and neutrons are also called

Accepted Answer

nucleons. Protons and neutrons are the particles found in the nucleus, so they are collectively called nucleons. The other choices describe different concepts: isotopes and isobars are types of atoms, and elements are pure substances defined by atomic number.

Question 2

From the $ \alpha $ -particle scattering experiment, Rutherford concluded that: This question has multiple correct options

Accepted Answer

the radius of the nucleus is less than $ 10^{-14} m $. Rutherford’s scattering showed that most of the atom is empty space and that the positive charge is concentrated in a tiny central nucleus. The existence of strong deflections at very small impact parameters implies the nucleus must have a radius smaller than about $10^{-14}\,\text{m}$.

Question 3

Which of following was not explained by Rutherford?

Accepted Answer

Arrangement of nuetrons. Rutherford explained the nuclear model of the atom and the general placement of electrons around a central nucleus. Neutrons were not yet discovered in his era, so their arrangement was not explained by him.

Question 4

The energy of an electron in the ground state of $ \boldsymbol{H} $ atom is $ -\mathbf{1 3 . 6 e V} $ The negative sign indicates that:

Accepted Answer

energy of the electron in the H atom is lower than that of a free electron. The reference level is usually taken as a free electron and proton infinitely far apart, with energy 0. A negative value means the electron is bound in the atom and has less energy than a free electron.

Question 5

Outer electronic configuration of Cl⁻ is:

Accepted Answer

3s²3p³3p³3p². The chloride ion (Cl⁻) has gained one electron compared to a neutral chlorine atom, resulting in a configuration of 3s²3p⁶, which corresponds to option D.

Question 6

The sub-shell with lowest value of (n + l) is filled up first. When two or more sub-shells have the same (n + l) value, the sub-shell with lowest value of 'n' is filled up first. Therefore, the correct order is:

Accepted Answer

4s 3d 4p 5s 4d. The correct order of filling sub-shells is determined by the (n + l) rule. Among sub-shells with the same (n + l) value, the one with the lower 'n' is filled first. Option A correctly represents the filling order: 4s (n+l=4), 3d (n+l=5), 4p (n+l=5), 5s (n+l=5), 4d (n+l=6).

Question 7

The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following?

Accepted Answer

d-electrons in Fe (Z = 26). Fe²⁺ has 6 d-electrons (26 total electrons minus 2 from the 4s orbital). Neutral Fe has 6 d-electrons as well, so the number of d-electrons in Fe²⁺ is equal to that in Fe. Thus, option B is incorrect.

Question 8

Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?

Accepted Answer

3s, 3p, 4s, 3d. The energy of orbitals increases as per the (n+l) rule, where n is the principal quantum number and l is the azimuthal quantum number. For titanium, the correct order is 3s < 3p < 4s < 3d.

Question 9

When a radioactive element emits successively one α-particle and two β-particles, the mass number of the daughter element:

Accepted Answer

is reduced by 4 units. Emission of an α-particle reduces the mass number by 4, while β-particle emissions do not affect the mass number. Thus, the mass number decreases by 4 units overall.

Question 10

If species ᵇₐX emits firstly a positron, then two α and two β and in last one α and finally converted to species ᶜₐY, so correct relation is:

Accepted Answer

c = a - 5, d = b - 8. The positron emission decreases the atomic number by 1, each α-particle emission decreases the atomic number by 2 and mass number by 4, and β-particle emission does not affect the mass number but increases the atomic number by 1. After applying these changes step-by-step, the final relation is c = a - 5 and d = b - 8.

Question 11

Number of neutrons in a parent nucleus X, which gives ¹⁴₇N nucleus, after two successive β emissions, would be:

Accepted Answer

9. The parent nucleus undergoes two successive β emissions to form ¹⁴₇N. Each β emission increases the atomic number by 1, so the parent nucleus must have been ¹⁴₅B. The number of neutrons in ¹⁴₅B is 14 (mass number) - 5 (atomic number) = 9.

Question 12

In a radioactive decay, an emitted electron comes from

Accepted Answer

The nucleus of atom. In beta decay, an electron is emitted from the nucleus when a neutron converts into a proton and an electron. This electron is not from the atomic orbitals but is produced within the nucleus.

Question 13

If an isotope of hydrogen has two neutrons in its atom, its atomic number and atomic mass number will respectively be

Accepted Answer

1 and 3. The atomic number of hydrogen is always 1, as it has one proton. If the isotope has two neutrons, its mass number will be 1 (proton) + 2 (neutrons) = 3.

Question 14

Emission of an alpha particle leads to a

Accepted Answer

Decrease of 2 units in the charge of the atom. Emission of an alpha particle (which is a helium nucleus) decreases the atomic number by 2 (charge of the atom) and the mass number by 4. Thus, the charge decreases by 2 units.

Question 15

According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h = 6.62 × 10⁻²⁷ ergs, c = 3 × 10¹⁰ cm s⁻¹, Nₐ = 6.02 × 10²³ mol⁻¹)

Accepted Answer

1.196 × 10⁸ λ. The energy absorbed per mole is calculated using the formula E = (h × c × Nₐ) / λ. Substituting the given values, the result matches option B.

Question 16

The energies E₁ and E₂ of two radiations are 25 eV and 50 eV, respectively. The relation between their wavelengths i.e., λ₁ and λ₂ will be:

Accepted Answer

λ₁ = 2λ₂. Energy (E) and wavelength (λ) are inversely proportional (E = hc/λ). Since E₂ is double E₁, λ₁ will be twice λ₂, i.e., λ₁ = 2λ₂.

Question 17

The value of Planck’s constant is 6.63 × 10⁻³⁴ Js. The velocity of light is 3.0 × 10⁸ m s⁻¹. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 × 10¹⁵ s⁻¹?

Accepted Answer

4 × 10¹. The wavelength (λ) is calculated using the formula λ = c / ν, where c is the speed of light and ν is the frequency. Substituting c = 3.0 × 10⁸ m/s and ν = 8 × 10¹⁵ s⁻¹, we get λ = (3.0 × 10⁸) / (8 × 10¹⁵) = 3.75 × 10⁻⁸ m = 37.5 nm, which is closest to 4 × 10¹ nm.

Question 18

Calculate the energy in joule corresponding to light of wavelength 485 nm: (Planck’s constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 10⁸ ms⁻¹)

Accepted Answer

4.42 × 10⁻¹⁸. The energy of light is calculated using the formula E = hc/λ. Substituting h = 6.63 × 10⁻³⁴ Js, c = 3 × 10⁸ m/s, and λ = 485 nm = 485 × 10⁻⁹ m, we get E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (485 × 10⁻⁹) = 4.42 × 10⁻¹⁸ J.

Question 19

Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region?

Accepted Answer

Balmer series. The Balmer series corresponds to electronic transitions where the final energy level is n=2, and it falls in the visible region of the electromagnetic spectrum.

Question 20

Based on equation E = -2.178 × 10⁻¹⁸ [2/n²] J, certain conclusions are written. Which of them is not correct?

Accepted Answer

For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.. Option C is incorrect because a more negative energy for n = 1 indicates that the electron is more tightly bound to the nucleus, not more loosely bound.

NEET Chemistry: Structure of Atom questions with solutions

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