JEE Advanced (Single Correct Type): Atoms & Nuclei | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The ratio of the speed of electrons in the ground state of hydrogen to the speed of light in vacuum is
(a) 1/2
(b) 2/137
(c) 1/137
(d) 1/237

Correct Answer is option (c)
Speed of electron in n^th orbit of hydrogen atom v =

In ground state

Q.2. A sodium atom is in one of the states labeled 'Lowest excited levels'. It remains in that state for an average time of 10^–8 sec, before it makes a transition back to a ground state. What is the uncertainty in energy of that excited state?
(a) 6.56 × 10^-8 eV 
(b) 2 × 10^-8 eV 
(c) 10^-8 eV
(d) 8 × 10^-8 eV

Correct Answer is option (a)
The average time that the atom spends in this excited state is equal to Δt, so by using ΔE.Δt = h/2π
⇒ Uncertainty in energy

Q.3. An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is
(a) 79.0
(b) 51.8
(c) 49.2
(d) 38.2

Correct Answer is option (a)
After the removal of first electron remaining atom will be hydrogen like atom.
So energy required to remove second electron from the atom

Total energy required = 24.6 + 54.4 = 79 eV.

Q.4. Hydrogen (H), deuterium (D), singly ionized helium (He⁺) and doubly ionized lithium (Li) all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of emitted radiations are λ₁, λ₂, λ₃ and λ₄ respectively. Then approximately
(a) λ₁ = λ₂ = 4λ₃ =9λ₄
(b) 4λ₁ = 2λ₂ = 2λ₃ = λ₄
(c) λ₁ = 2λ₂ = 2√2λ₃ = 3√2λ₄
(d) λ₁ = λ₂ = 2λ₃ = 3√2λ₄

Correct Answer is option (a)
Using ΔE ∝ Z²   (∵ n₁ and n₂ are same)

Q.5. The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of 3
(a) 10²⁰
(b) 10¹⁹
(c) 10¹⁷
(d) 10¹⁵

Correct Answer is option (d)
mvr = h/2π (for first orbit)

Q.6. Consider an electron (m = 9.1 × 10⁻³¹ kg) confined by electrical forces to move between two rigid walls separated by 1.0 × 10⁻⁹ metre, which is about five atomic diameter. The quantised energy value for the lowest stationary state is:
(a) 12 × 10⁻²⁰J
(b) 6.0 × 10⁻²⁰J
(c) 6.0 × 10⁻¹⁸J
(d) 6J

Correct Answer is option (b)
It will form a stationary wave
λ = 2ℓ = 2 × 10⁻⁹m

Q.7. In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy E₀ and the radius r₀ of the first orbit will be (a₀ is the Bohr radius)
(a) E₀ = -27.2 eV; r₀ = a₀/2
(b) E₀ = -27.2 eV; r₀ = a₀
(c) E₀ = -13.6 eV; r₀ = a₀/2
(d) E₀ = -13.6 eV; r₀ = a₀

Correct Answer is option (a)
Here radius of electron orbit r ∝ 1/m and energy E ∝ m,
where m is the mass of the electron.
Hence energy of hypothetical atom
E₀ = 2 × (-13.6 eV) = -27.2 eV eV and radius r₀ = a₀/2

Q.8. A neutron with velocity V strikes a stationary deuterium atom, its kinetic energy changes by a factor of
(a) 15/16
(b) 1/2
(c) 2/1
(d) None of these

Correct Answer is option (d)
Neutron velocity ¾ v, mass ¾ m
Deuteron contains 1 neutron and 1 proton, mass ¾ 2m

In elastic collision both momentum and K.E. are conserved Pi¾ pf
mv = m₁v₂ + m₂v₂ ⇒ mv = mv₁ + 2mv₂ ... (i)
By conservation of kinetic energy
.. (ii)
By solving (i) and (ii) we get

(Fractional change in K.E.)

Q.9. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr's atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength l (given in terms of the Rydberg constant R for the hydrogen atom) equal to
(a) 9 / (5R) 
(b) 36 / (5R) 
(c) 18 / (5R) 
(d) 4 / R

Correct Answer is option (c)
In hydrogen atom E_n = - Rhc/n²
 Also E_n µm ; where m is the mass of the electron.
Here the electron has been replaced by a particle whose mass is double of an electron.
Therefore, for this hypothetical atom energy in n^th orbit will be given by

The longest wavelength l_max (or minimum energy) photon will correspond to the transition of particle from n =3 to

This gives

Q.10. The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is 1.4 kilowatt/m². The average earth-sun distance is 1.5 × 10¹¹ m . The mass lost by the sun per day is (1 day ¾ 864 seconds)
(a) 4.4 × 10⁹ kg
(b) 7.6 × 10¹⁴ kg
(c) 3.8 × 10¹² kg

Correct Answer is option (d)
Energy radiated ¾ 1.4 kW/m²


Total energy radiated/day