$$^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He}$$
Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^{4}\text{He}$ as $238.050\text{ u}$, $234.043\text{ u}$ and $4.003\text{ u}$, respectively. The Q value for the reaction, in keV, is :
[Given : $1\text{ u} = 931.5\text{ MeV c}^{-2}$]
Nuclear Reaction Q-Value Calculation
The Q-value of a nuclear reaction is the energy released or absorbed. It is determined by the difference in mass between the reactants and the products.
Mass Defect Determination
The given nuclear reaction is:
$ ^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He} $The masses provided are:
- Mass of reactant $^{238}\text{U}$: $m_{\text{reactants}} = 238.050 \text{ u}$
- Mass of products $^{234}\text{Th} + ^{4}\text{He}$: $m_{\text{products}} = 234.043 \text{ u} + 4.003 \text{ u} = 238.046 \text{ u}$
The mass defect ($\Delta m$) is calculated as:
$ \Delta m = m_{\text{reactants}} - m_{\text{products}} $ $ \Delta m = 238.050 \text{ u} - 238.046 \text{ u} $ $ \Delta m = 0.004 \text{ u} $Q-Value Calculation
The Q-value is obtained by converting the mass defect into energy using the conversion factor $1 \text{ u} = 931.5 \text{ MeV/c}^2$.
$ Q = \Delta m \times 931.5 \text{ MeV/u} $ $ Q = 0.004 \text{ u} \times 931.5 \text{ MeV/u} $ $ Q = 3.726 \text{ MeV} $To express the Q-value in keV, we use the conversion $1 \text{ MeV} = 1000 \text{ keV}$.
$ Q = 3.726 \times 1000 \text{ keV} $ $ Q = 3726 \text{ keV} $Therefore, the Q-value for the reaction is 3726 keV.
Similar Questions
In the first excited state of hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately :
(Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}, \text{ e} = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)Four statements are given (A is mass number) :
A. The volume of a nucleus is proportional to $A^{1/3}$.
B. The volume of a nucleus is proportional to A.
C. The difference in mass of an atom and its nucleus is called the mass defect.
D. The difference in mass of a nucleus and its constituent nucleons is called the mass defect.
Choose the correct answer from the options given below :- An unknown nucleus has a nuclear density of $2.29 \times 10^{17} \text{ kg/m}^3$ and mass of $19.926 \times 10^{-27} \text{ kg}$. Its mass number A is approximately :
(Take $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$) - In Geiger-Marsden experiment, the number of scattered $\alpha$-particles $N(\theta)$ is plotted as a function of scattering angle $\theta$. Which of the following options represents the correct plot ?
- A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$, then variation of the photocurrent $I$ and magnitude of the stopping potential $|V|$ with time is best represented by :
- A ray of light with wavelength $\lambda$ is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelength of these photo-electric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 \gg \lambda$. The correct option is :
An ideal Zener diode with breakdown voltage of $-3\text{ V}$ is reverse biased with a negative input voltage $V_i = -5\text{ V}$. The magnitude of voltage difference between points B and A is :
- A photon and an electron, each of $20\text{ eV}$ energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{ph}$, $\frac{p_e}{p_{ph}}$ is :
(Take speed of light $= 3 \times 10^8\text{ ms}^{-1}$, charge of electron $= -1.6 \times 10^{-19}\text{ C}$ and mass of electron $= 9 \times 10^{-31}\text{ kg}$) - Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5\text{ ms}^{-1}$. The radius of the orbit is $x \times 10^{-9}\text{ m}$. The value of $x$ is :
[Take the mass of electron to be $9 \times 10^{-31}\text{ kg}$, charge of electron $= 1.6 \times 10^{-19}\text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$] - For a metal of work function $6.6 \text{ eV}$, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect?
(Take Planck's constant as $6.6 \times 10^{-34} \text{ J s}$)
Important Questions from Modern Physics
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
- A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$) - The energy released when $\frac{7}{17.13} \text{ kg}$ of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is _________. (Nearest integer)
(Mass of ${}^7_3\text{Li} = 7.0183 \text{ u}$, mass of ${}^4_2\text{He} = 4.004 \text{ u}$, mass of proton = $1.008 \text{ u}$ and $1 \text{ u} = 931 \text{ MeV/}c^2$ and Avogadro number $= 6.0 \times 10^{23}$) - A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy $3.2$ Mev. The estimation of the radius of the nucleus of gold element $(Z = 79)$ by him is $\times 10^{-16}$ m
(Take $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$)
