A radioactive element forms its own isotope after 3 consecutive disintegrations. The particles emitted are

3 β–particles
2 β–particles and 1 α–particle
3 β–particles and 1 α–particle
2 β–particles and 1 α–particle

Solution

The electrons cannot exist inside the nucleus because

de-Broglie wavelength associated with electron in β-decay is much less than the size of nucleus
de-Broglie wavelength associated with electron in β-decay is much greater than the size of nucleus
de-Broglie wavelength associated with electron in β-decay is equal to the size of nucleus
negative charge cannot exist in the nucleus

A radioactive nuclide is produced at the constant rate of n per second (say, by bombarding a target with neutrons).The expected number N of nuclei in existence t seconds after the number is N₀ is given by

Where λ is the decay constant of the sample

Solution

At time t = 0, N₁ nuclei of decay constant λ₁ and N₂ nuclei of decay constant λ₂ are mixed. The decay rate of mixture is

\(-N_{1}N_{2}o-(\lambda _{1}+\lambda _{2})t\)
\(-\left ( \frac{N_{1}}{N_{2}} \right )e-(\lambda _{1}+\lambda _{2})t\)
\(-(N_{1}\lambda _{1}e^{-\lambda _{1}^{t}}+N_{1}\lambda _{2}e^{-\lambda _{2}^{t}})\)
\(-N_{1}\lambda _{1}N_{2}\lambda _{2}e^{(-\lambda _{1}+\lambda _{2})t}\)

If the end A of a wire is irradiated with α-rays and the other end B is irradiated with β-rays.Then

a current will flow from A to B
a current will flow from B to A
there will be no current in the wire
a current will flow from each end to the mid-point of the wire

Solution

Due to irradiation of α-rays on end A will make it(positive) and irradiation of β-rays on end B will make it (negative) hence current will flow from A to B(or from positive to negative).

Radium ²²⁶Ra, spontaneously decays to radon with the emission of an α-particle and a γ ray. If the speed of the α particle upon emission from an initially stationary radium nucleus is 1.5 × 10⁷m/s, what is the recoil speed of the resultant radon nucleus? Assume the momentum of γ ray is negligible compared to that of α particle.

2..0 × 10⁵m/s
2.7 × 10⁵m/s
3.5 × 10⁵m/s
1.5 × 10⁷m/s

Solution

If an electron and positron annihilate, then the energy released is

3.2 × 10^-13 J
1.6 × 10^-13 J
4.8 × 10^-13 J
6.4 × 10^-13 J

Solution

A radioactive nucleus undergoes α-emission to form a stable element. What will be the recoil velocity of the daughter nucleus if v is the velocity of α emission?

\(\frac{4v}{A-4}\)
\(\frac{2v}{A-4}\)
\(\frac{4v}{A+4}\)
\(\frac{2v}{A+4}\)

Solution

Radioactive element decays to form a stable nuclide,then the rate of decay of reactant is

Solution

In the uranium radioactive series, the initial nucleus is ₉₂U²³⁸ and that the final nucleus is ₈₂Pb²⁰⁶. When uranium nucleus decays to lead, the number of α particles and β particles emitted are

8α, 6β
6α, 7β
6α, 8β
4α, 3β

Solution

Let no.of α-particles emitted be x and no. of β particles emitted be y.

Diff. in mass no. 4x = 238 – 206 = 32 ⇒ x = 8

Diff.in charge no. 2x – 1y = 92 – 82 =10

16 – y =10, y = 6

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