Consider a beta decay reaction:{}_1^3H to {}_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The $Q$-value of the $\beta$ decay reaction is given by the mass difference between the parent nucleus and the products.Since the atomic mass of $

Vedclass · Accepted Answer

$0.0186$

Vedclass · Answer

(C) The $Q$-value of the $\beta$ decay reaction is given by the mass difference between the parent nucleus and the products.Since the atomic mass of ${}_1^3H$ includes one electron and the atomic mass of ${}_2^3He$ includes two electrons,we must account for the electron emitted in the decay.The $Q$-value is calculated as:$Q = [M({}_1^3H) - M({}_2^3He)] 	imes c^2$Note: The mass of the emitted electron is already accounted for in the atomic mass difference of the neutral atoms.$Q = (3.016050 \, u - 3.016030 \, u) 	imes 931.5 \, MeV/u$$Q = 0.000020 \, u 	imes 931.5 \, MeV/u$$Q = 0.01863 \, MeV$The maximum kinetic energy of the electron is approximately equal to the $Q$-value of the reaction.Therefore,the maximum energy is $0.0186 \, MeV$.

List-$I$	List-$II$
$(P)$ ${ }_{92}^{238} U \rightarrow{ }_{91}^{234} Pa$	$(1)$ $1 \alpha$ and $1 \beta^{+}$
$(Q)$ ${ }_{82}^{214} Pb \rightarrow{ }_{82}^{210} Pb$	$(2)$ $3 \beta^{-}$ and $1 \alpha$
$(R)$ ${ }_{81}^{210} Tl \rightarrow{ }_{82}^{206} Pb$	$(3)$ $2 \beta^{-}$ and $1 \alpha$
$(S)$ ${ }_{91}^{228} Pa \rightarrow{ }_{88}^{224} Ra$	$(4)$ $1 \alpha$ and $1 \beta^{-}$
	$(5)$ $1 \alpha$ and $2 \beta^{+}$

Consider a $\beta$ decay reaction:
${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$
The atomic masses of ${}_1^3H$ and ${}_2^3He$ are $3.016050 \, u$ and $3.016030 \, u$,respectively. Find the maximum possible energy of the electron in $MeV$.

Similar Questions

In the nuclear decay sequence given below:
$_Z{X^A} \to {}_{Z + 1}{Y^A} \to {}_{Z - 1}{K^{A - 4}} \to {}_{Z - 1}{K^{A - 4}}$
the particles emitted in the sequence are:

$A$ free neutron decays spontaneously into :

In the uranium series,the decay constant of the parent nuclide is $\lambda$. What is the decay constant of the stable end product of this series?

In $\beta^{-}$ decay,a neutron transforms into a proton within the nucleus according to the equation:
$\text{neutron} \rightarrow \text{proton} + \beta^{-} + x$
In this equation,the particle represented by '$x$' is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider a $\beta$ decay reaction:${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$The atomic masses of ${}_1^3H$ and ${}_2^3He$ are $3.016050 \, u$ and $3.016030 \, u$,respectively. Find the maximum possible energy of the electron in $MeV$.