Tag: nuclei

Quizzes (7) Questions (160)

Questions Related to nuclei

Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A positron is emitted by radioactive nucleus of proton number $90$. The product nucleus will have proton number :

  1. $91$
  2. $90$
  3. $89$
  4. $88$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given nuclear reaction is-

$ _{90}X \rightarrow   _{+1}e $  $+ $ $ _zY$ 
Using law of conservation of atomic (or proton) number:
$90 = 1 + Z$
$\implies$ $Z = 89$
Thus the product nucleus will have proton number $89$.

Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

When $ _{15}P^{30}$ decays to become  $ _{14}Si^{30}$, which particle is released ?

  1. electron
  2. $\alpha$-particle

  3. neutron

  4. positron
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The nuclear reaction :   $ _{15}P^{30}\rightarrow$   ${14}Si^{30} + $  $ _{+1}e^0$

Thus a positron is emitted during the decay of  $ _{15}P^{30}$ into   $ _{14}Si^{30}$.

Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A nucleus $ _{  }^{ 220 }{ X }$ at rest decays emitting an $\alpha$- particle. If energy of daughter nucleus is $0.2MeV$, $Q$ value of the reaction is

  1. $10.8MeV$
  2. $10.9MeV$
  3. $11MeV$
  4. $11.1MeV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a nucleus of mass number 220 at rest decays by emitting an alpha particle, momentum is conserved. The alpha particle (mass 4) and daughter nucleus (mass 216) recoil in opposite directions. The Q-value is the sum of kinetic energies of the alpha particle and the daughter nucleus. Using momentum conservation, K_alpha / K_daughter = M_daughter / M_alpha = 216 / 4 = 54. Given K_daughter = 0.2 MeV, K_alpha = 54 * 0.2 = 10.8 MeV. Thus, Q = K_alpha + K_daughter = 10.8 + 0.2 = 11 MeV.

Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Which word equation represents $\beta^+$ decay?

  1. Proton $\rightarrow$ neutron $+$ electron $+$ electron antineutrino
  2. Proton $\rightarrow$ neutron $+$ electron $+$ electron neutrino
  3. Proton $\rightarrow$ neutron $+$ positron $+$ electron antineutrino
  4. Proton $\rightarrow$ neutron $+$ positron $+$ electron neutrino
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 Positive beta decay $(β+\ \ \text{decay})$ also called the  positron emission. In this decay  a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus, and the nucleus emits a neutrino and a positron, which is a positive particle like an ordinary electron in mass but of opposite charge. Thus, positive beta decay produces a daughter nucleus, the atomic number of which is one less than its parent and the mass number of which is the same.

The equation can be written as :

$p \rightarrow n+ e^+ + v _e$

Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Which of the following nuclei is produced when a $ _{92}U^{238}$ nucleus undergoes a $(d, 2n)$ reaction followed by a beta decay?

  1. $ _{93}Np^{238}$
  2. $ _{94}Pu^{239}$
  3. $ _{94}Pu^{238}$
  4. $ _{92}U^{238}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A (d, 2n) reaction on U-238 (Z=92, A=238) involves adding a deuteron (Z=1, A=2) and removing two neutrons (Z=0, A=1 each), resulting in Z=93, A=238 (Np-238). A subsequent beta decay (Z increases by 1, A stays same) results in Z=94, A=238, which is Pu-238.

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

Which of the following relations is correct?

  1. $E = mc$
  2. $E = mc^2$
  3. $E = 2mc^2$
  4. $E = mc^2/4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Einstein, energy and mass are related by the relation.
$E = mc^2$
where c is the speed of light in vacuum.

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

One milligram of matter is converted into energy. The energy released will be

  1. $9\times 10^{6} J$
  2. $9\times 10^{8}J$
  3. $9\times 10^{10}J$
  4. $9\times 10^{12}J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $m = 1\ mg = 1\times 10^{-3} g $

             $= 1\times 10^{-6}kg$
According to Einstein mass-energy equivalence
                $E = mc^{2}$
where $c$ is the speed of light in vacuum
$\therefore E = (1\times 10^{-6} kg)(3\times 10^{8}ms^{-1})^{2}$
$ = 9\times 10^{10}J$.

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

The relation between the volume $V$ and the mass $M$ of a nucleus is:

  1. $V\propto M^{3}$
  2. $V\propto M^{1/3}$
  3. $V\propto M$
  4. $V\propto 1/M$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the density of nucleus is fixed.

$D=\cfrac { M }{ V } \ \Rightarrow M=DV\ M\propto V$