$$^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He}$$
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 $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)- 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 $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$) 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.
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(Take Planck's constant as $6.6 \times 10^{-34} \text{ J s}$) Match List I with List II :
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(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 $= 9 \times 10^{-31}\text{ kg}$)
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 ______ $\Omega$.
The binding energy for the following nuclear reactions are expressed in MeV.
${}_2\text{He}^3 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^4 + 20 \text{ MeV}$
${}_2\text{He}^4 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^5 - 0.9 \text{ MeV}$
If $\text{X}_3, \text{X}_4, \text{X}_5$ denote the stability of ${}_2\text{He}^3, {}_2\text{He}^4$ and ${}_2\text{He}^5$, respectively, then the correct order is :Identify the correct truth table of the given logic circuit.
- When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is $3.2 \text{ V}$. If a second light having wavelength twice of first light is used, the stopping potential drops to $0.7 \text{ V}$. The wavelength of first light is _______ m.
($h = 6.63 \times 10^{-34} \text{ J.s}, e = 1.6 \times 10^{-19} \text{ C}, c = 3 \times 10^8 \text{ m/s}$) - 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 $= 6.6 \times 10^{-34}\text{ J.s}$)
