Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

The particle emitted in the nuclear reaction 
$ _{z}\textrm{X}^{A}$ = $ _{z+1}\textrm{Y}^{A}$ + ..... will be 

  1. $\alpha$ -particle
  2. $\beta^{-}$ -particle
  3. $\beta{+}$ -particle
  4. $Photon$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In $\beta^{-}$ decay, there is a gain of one proton and an anti-neutrino is emitted from nucleus, hence atomic number is increased by 1 and weight remains same.
Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The nucleus of mass  $M +  \Delta m$  is at rest and decays into two daughter nuclei of equal mass  $\dfrac {  M  } { 2 }$  each. Speed of light is  $ c.$ The speed of daughter nuclei is

  1. $c \dfrac { \Delta m } { M + \Delta m }$
  2. $c \sqrt { \dfrac { 2 \Delta m } { M } }$
  3. $c \sqrt { \dfrac { \Delta m } { M } }$
  4. $c \sqrt { \dfrac { \Delta m } { M + \Delta m } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy released Q = delta_m * c^2. This energy is shared as kinetic energy between the two daughter nuclei. K1 + K2 = Q. Since masses are equal, velocities are equal. 2 * (1/2 * (M/2) * v^2) = delta_m * c^2. (M/2) * v^2 = delta_m * c^2. v^2 = (2 * delta_m / M) * c^2. v = c * sqrt(2 * delta_m / M).

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

The particle $X$ in the following nuclear reaction is  $ _{7}^{13}\textrm{N}$ $\longrightarrow $ $ _{6}^{13}\textrm{C}+$  $ _{1}^{0}\textrm{e}$ + $X$

  1. $P$
  2. $v$
  3. $e^{-}$
  4. $\alpha$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given reaction :      $^{13} _7N\rightarrow ^{13} _6C + ^0 _1e + ^A _Z X$

Conservation of mass number :      $13 = 13 + 0 +A$              $\implies A=0$
Conservation of atomic number :      $7 = 6 + 1 +Z$              $\implies Z=0$
Thus, the particle $X$ is neutrino i.e.  $\nu$

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

A radioactive material initially contains $10gm$ and after few days $3gm$ is left, then the emission rate of $\alpha$ or $\beta$ particle:-

  1. Will continue as usual
  2. Becomes $0.3$times

  3. Increases

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The emission rate of a radioactive material is proportional to the number of radioactive nuclei present (activity A = lambda * N). Since the material decreased from 10g to 3g, the amount of radioactive material is now 0.3 times the initial amount. Therefore, the emission rate also becomes 0.3 times the original rate.

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

The ionization potential for second He electron is 

  1. 13.6 V

  2. 27.2 V

  3. 54.4 V

  4. 3.4 V
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ionization energy of a hydrogen-like ion is given by E = 13.6 * Z^2 / n^2 eV. For the second electron of Helium (He+), Z = 2 and n = 1, so the energy is 13.6 * (2^2) / 1^2 = 13.6 * 4 = 54.4 eV.

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

A radioactive substance contains a number of identical nuclei that emit $\beta$- particles. Which property of these nuclei remains unaltered by emission?

  1. charge

  2. neutron number

  3. nucleon number

  4. proton number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that in any $\beta -$ emission the mass number of the nucleus does not changes .

This implies that the nucleon number will remain same. 
So, option C is correct.

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

In $\beta^-$ decay, a

  1. neutron converts into a proton emitting antineutrino.

  2. neutron converts into a proton emitting neutrino.

  3. proton converts into a neutron emitting antineutrino.

  4. proton converts into a neutron emitting neutrino.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The reaction of beta decay is shown as :

$n \rightarrow p + e^- + \bar{\nu} _e$               (antineutrino)

So, in a $\beta-$ decay, a neutron converts into a proton emitting antineutrino.

option (A) is correct.

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.