Test: Atoms and Nuclei - 1 - CUET Humanities MCQ

When a charged particle is accelerated by an electric field (uniform or non-uniform), by work energy theorem, ΔKE = W, we have
 mv² -  mu² = qV (∵ W = qV)
For electron,
 mv² -  mu² = eV
or v = 
If electron is initially at rest, u = 0
v = 

Radius of orbit = 
Where, n = 1, Z = 1
So, r = 0.53 Å

The kinetic and potential energies of an electron in the n^th excited state are given by

Adding (i) and (ii), we get the total energy E which is

Notice from (i) and (iii) that E = -KE. Given E = -3.4 eV.
Hence KE = -E = -(-3.4) = +3.4 eV.
Thus the correct choice is (2).

Moment of momentum is angular momentum.
According to Bohr's model of atom, electrons can revolve only in those orbits in which their angular momentum is an integral multiple of h/2π, where h is Planck's universal constant.
If J is moment of momentum, i.e. angular momentum, then from Bohr's postulate, we have:
J = nh/2π, where n is an integer (1, 2, 3…), called the principal quantum number of the orbit
In the second orbit, n = 2
Therefore,
J = 2h/2π = h/π

The radioactivity of a sample decreases to (1/2ⁿ) in 'n' half lives. Since the initial activity is 64 times the permissible level, it must become 1/64^th of its initial value.
∴ 1/64 = 1/2ⁿ or n = 6
In other words, in six half-lives, the activity will reduce itself to the permissible level. Since half-life is 2 hours, the total time taken = 2 × 6 h = 12 hours.

Binding Energy of ₁H¹ = 2.22/2 = 7.08
Binding Energy of ₂He⁴ = 28.3/4 = 7.08
Binding Energy of ₂₆Fe⁵⁶ = 492/56 = 8.78
Binding Energy of ₉₂U²³⁵ = 1786/235 = 7.6
As the binding energy of Fe⁵⁶ is highest, so the most stable nucleus will be ₂₆Fe⁵⁶.

The mass number reduces by 200 - 168 = 32.
Hence, 8 α-particles are emitted. The emission of 8 α-particles reduces the atomic number by 16. But the atomic number reduces by 90 - 80 = 10.
Hence, the number of β-particles emitted = 16 - 10 = 6.
Hence, option (1) is correct.

Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a β-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from a higher to a lower energy state.
In negative β-decay, a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted β-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

After one α-emission, the daughter nucleus reduces in mass number by 4 units and in atomic number by 2 units.
In β-emission, the atomic number of daughter nucleus increases by 1 unit.
The reaction can be written as:

Thus, the resulting nucleus is:
₈₂Y²⁰⁶, i.e. ₈₂Pb²⁰⁶

The binding energy per nucleon for the middle nuclides (from A = 20 to A = 56) is maximum.
Hence, these are more stable.
As the mass number increases, the binding energy per nucleon gradually decreases and ultimately, the binding energy per nucleon of heavy nuclides (such as uranium. etc.) is comparatively low. Hence, these nuclides are relatively unstable. So, they can be fissioned easily.