Download App

Physics 2 Marks Questions

The Nucleus · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 11 questions.

Questions

2 Marks Questions

Take a timed test

11 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
A free neutron beta-decays to a proton with a half-life of 14 minutes.
  1. What is the decay constant?
  2. Find the energy liberated in the process.
Answer
$\text{In}\rightarrow\text{P + e}^-$
We know : Half life $=\frac{0.6931}{\lambda}$ (Where $\lambda$ = decay constant).
$\lambda=\frac{0.6931}{14\times60}=8.25\times10^{-4}\text{S}$ [As half life = 14min = 14 × 60sec].
Energy = [Mn - (MP + Me)]u = [(Mnu - Mpu) - Mpu]c2 = [0.00189u - 511KeV/c2]
= [1293159 ev/c2 - 511000ev/c2]c2 = 782159eV = 782Kev.
View full question & answer→
Question 22 Marks
Lithium (Z = 3) has two stable isotopes 6Li and 7Li. When neutrons are bombarded on lithium sample, electrons and $\alpha$-particles are ejected. Write down the nuclear process taking place.
Answer
$\text{ }^6_3\text{Li + n}\rightarrow\text{ }^7_3\text{Li};\text{ }^7_3\text{Li + r}\rightarrow\text{ }^8_3\text{Li}$
$\text{ }^8_3\text{Li}\rightarrow\text{ }^8_4\text{Be + e}^-+\text{v}^-$
$\text{ }^8_4\text{Be}\rightarrow\text{ }^4_2\text{He}+\text{ }^4_2\text{He}$
View full question & answer→
Question 32 Marks
In the decay 64Cu → 64Ni + e+ + v, the maximum kinetic energy carried by the positron is found to be 0.650MeV.
  1. What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150MeV?
  2. What is the momentum of this neutrino in kg-m/s?
Use the formula applicable to a photon.
Answer
64Cu → 64Ni + e+ + v,

Emission of nutrino is along with a positron emission.

  1. Energy of positron = 0.650MeV.

Energy of Nutrino = 0.650 - KE of given position = 0.650 - 0.150 = 0.5MeV = 500Kev.

  1. Momentum of Nutrino $=\frac{500\times1.6\times10^{-19}}{3\times10^8}\times10^3\text{J}=2.67\times10^{-22}\text{kg m/s}.$
View full question & answer→
Question 42 Marks
Radioactive 131I has a half-life of 8.0 days. A sample containing 131I has activity $20\mu\text{Ci}$ at t = 0.
  1. What is its activity at t = 4 days?
  2. What is its decay constant at t = 4.0 days?
Answer
$\text{t}_{\frac{1}{2}}=8.0\text{ days};\text{ A}_0=20\mu\text{Cl}$

  1. $\text{t}=4.0\text{ dys};\lambda=\frac{0.693}{8}$

$\text{A = A}_{0}\text{e}^{-\lambda\text{t}}=20\times10^{-6}\times\text{e}^{\big(\frac{-0.693}{8}\big)\times4}$

$=1.41\times10^{-5}\text{Ci}=14\mu\text{Ci}$

  1. $\lambda=\frac{0.693}{8\times24\times3600}=1.0026\times10^{-6}.$
View full question & answer→
Question 52 Marks
The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old $\text{ }^{32}\text{P}\big(\text{t}_{\frac{1}{2}}=14.3\text{days}\big)$ source if it was originally purchased for 800 rupees?
Answer
$\text{t}_{\frac{1}2{}}=14.3\text{ days};\text{ t}=30\text{ days}=1\text{ month}$
As, the selling rate is decided by the activity, hence A0 = 800 disintegration/sec.
We know, $\text{A = A}_0\text{e}^{-\lambda\text{t}}$ $\Big[\lambda=\frac{0.693}{14.3}\Big]$
$\text{A}=800 \times 0.233669 = 186.935 = 187\text{ rupees.}$
View full question & answer→
Question 62 Marks
57Co decays to 57Fe by $\beta^+-\text{ emission.}$- emission. The resulting 57Fe is in its excited state and comes to the ground state by emitting $\gamma-\text{rays}.$ The half-life of $\beta^+-\text{decay}$ is 270 days and that of the $\gamma-\text{emissions}$ is 10-8 s. A sample of 57Co gives 5.0 × 109 gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5 × 109per second?
Answer
According to the question, the emission rate of $\gamma-\text{rays}$ will drop to half when the $\beta^+\text{decays}$ to half of its original amount. And for this the sample would take 270 days.
$\therefore$ The required time is 270 days.
View full question & answer→
Question 72 Marks
Calculate the mass of an $\alpha$-particle. Its binding energy is 28.2MeV.
Answer
Let the mass of $'\alpha'$ particle be xu.
$'\alpha'$ particle contains 2 protons and 2 neutrons.
$\therefore$ Binding energy = (2 × 1.007825u × 1 × 1.00866u - xu)C2 = 28.2MeV (given).
$\therefore$ x = 4.0016u.
View full question & answer→
Question 82 Marks
When a boron nucleus $\big(\text{ }^{10}_5\text{B}\big)$ is bombarded by a neutron, an $\alpha$-particle is emitted. Which nucleus will be formed as a result?
Answer
It is given that when a boron nucleus$\big(\text{ }^{10}_5\text{B}\big)$ is bombarded by a neutron, an $\alpha$-particle is emitted.
Let X nucleus be formed as a result of the bombardment.
According to the charge and mass conservation,
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ }^{10}_5\text{B}+^1_0\text{n}\rightarrow\text{X}+^4_2\text{He}\\\text{Charge}: \ \ 5 \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ 2\\\text{Mass}:\ \ \ \ 10 \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ 7 \ \ \ \ \ \ \ \ \ 4$
The mass number of X should be 7 and its atomic number should be 3.
$\therefore\text{X}=^7_3\text{Li}$
View full question & answer→
Question 92 Marks
The masses of 11C and 11B are respectively 11.0114u and 11.0093u. Find the maximum energy a positron can have in the $\beta^+$-decay of 11C to 11B.
Answer
Given:
Mass of 11C, m(11C) = 11.0114u
Mass of 11B, m(11B) = 11.0093u
Energy liberated in the $\beta^+$ decay (Q) is given by
Q = [m(11C) - m(11B) - 2me]c2
= (11.0114u - 11.0093u - 2 × 0.0005486u)c2
= 0.0010028 × 931MeV
= 0.9336MeV = 933.6keV
For maximum KE of the positron, energy of neutrino can be taken as zero.
$\therefore$ Maximum KE of the positron = 933.6keV
View full question & answer→
Question 102 Marks
The decay constant of $\text{ }^{197}_{80}\text{Hg}$ (electron capture to $\text{ }^{197}_{79}\text{Au}$) is 1.8 × 10-4 S-1.
  1. What is the half-life?
  2. What is the average-life?
  3. How much time will it take to convert 25% of this isotope of mercury into gold?
Answer
  1. $\text{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}$ [$\lambda\rightarrow$ Decay constant]

$\Rightarrow\text{t}_{\frac{1}{2}}=3820\text{sec}=64\text{min}$

  1. Average life $=\frac{\text{t}_{\frac{1}{2}}}{0.693}=92\text{min}.$
  2. $0.75=1\text{e}^{-\lambda\text{t}}\Rightarrow\text{In }0.75=-\lambda\text{t}$

$\Rightarrow\text{t}=\text{In}\frac{0.75}{-0.00018}=1598.23\text{sec.}$

View full question & answer→
Question 112 Marks
Complete the following decay schemes.
  1. $\text{ }^{226}_{88}\text{Ra}\rightarrow\alpha+$
  2. $\text{ }^{19}_8\text{O}\rightarrow\text{ }^{19}_9\text{F}+$
  3. $\text{ }^{25}_{13}\text{Al}\rightarrow\text{ }^{25}_{12}\text{Mg}+$
Answer

  1. $\text{ }^{226}_{88}\text{Ra}\rightarrow\text{ }^4_2\alpha+\text{ }^{222}_{26}\text{Rn}$

  2. $\text{ }^{19}_8\text{O}\rightarrow\text{ }^{19}_9\text{F}+\bar{\text{e}}+\bar{\text{v}}$

  3. $\text{ }^{25}_{13}\text{Al}\rightarrow\text{ }^{25}_{12}\text{Mg}+\text{e}^++\text{v}$

View full question & answer→
Generate Paper FreeAll The Nucleus topics
2 Marks Questions - Physics STD 12 Science Questions - Vidyadip