A nucleus with mass number \$220\$ initially at rest emits an \$\alpha \ particle\$. If the \$Q\$ value of the reaction is \$5.5 MeV\$, calculate the kinetic energy of the \$\alpha\ particle\$

\$ M \rightarrow X+ \alpha\$\$ E = \cfrac{p^2}{2m}\$Let Total energy = \$T\$\$ T = E_{\alpha} + E_{X}\$\$M = M_{\alpha} + M_{X}\$\$M_{X}= 216\$Conservation of momentum\$ p_{\alpha} = p_{X}\$\$ m_{\alpha} E_{alpha} = m_XE_X\$\$E_{\alpha} = \cfrac{m_x}{M}T\$\$ E_{\alpha} = \cfrac{216}{220} 5.5 = 5.4 MeV\$

A nucleus \$_n^m X\$ emits one \$\alpha\$ particle and two \$\beta\$ particles. The resulting nucleus is

In alpha decay, decreases 4 unit mass number and 2 unit atomic number.In beta decay, increases one unit atomic number but mass will remain same.So, \$_n^mX \rightarrow^{\alpha}\$ \$ _{n-2}^{m-4}Y \rightarrow ^{2\beta}\$ \$_{n-2+2}^{m-4}Z\$ or \$ _n^{m-4}Z\$

Calculate the energy released if \$U^{238}\$-nucleus emits an \$\alpha-\$particle.ORCalculate the energy released in MeV in the following nuclear reaction\$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He + Q\$Given Atomic mass of \$^{238}U = 238.05079 \,u\$Atomic mass of \$^{234}Th = 234.04363 \,u\$Atomic mass of alpha particle \$= 4.00260 \,u\$\$1 \,u = 931.5 \,MeV / c^2\$Is the decay spontaneous ? Give reasons.

The process is \$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He + Q\$The energy released \$(\alpha - particle)\$\$Q = (M_U - M_{TH} - M_{He}) c^2\$\$= (238.05079 - 234.04363 - 4.00260) u \times c^2 = (0.00456 \,u) \times c^2\$\$= 0.00456 \times \left ( \dfrac{931. - MeV}{c^2} \right ). c^2 = 4.25 \,MeV\$Yes, the decay is spontaneous (since Q is positive).

A radioactive nucleus (initial mass number A and atomic number Z) emits \$3\$ \$\alpha\$-particles and \$2\$ positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

When a radioactive nucleus emits 1 α-particle mass number decreases by 4 and the atomic number decreases by 2.So after the emission of \$3 - \alpha\$ particlesNew atomic mass, \$A' = A - 4 \times 3 = A - 12\$ New atomic number \$Z'= Z - 6\$When this nucleus emits \$1 β\$-particle (positron), the atomic mass remains unchanged but the atomic number decreases by 1.So after emission of 2 positrons \$A" = A' = A-12\$ \$Z" = Z'-2×1 = Z-8\$So, the ratio of neutrons to protons-\$\dfrac{n}{p} = \dfrac{A-Z-4}{Z-8}\$

A nucleus with \$Z=92\$ emits the following in a sequence: \$\alpha, \alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha ; \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha\$. The \$Z\$ of the resulting nucleus is

For each \$\alpha\$ emission z decreases by 2; For each \$\beta^{+}\$ emission, z decreases by 1 and that for each \$\beta^{-}\$ it increases by 1.No of \$\alpha\$ emitted \$= 8 \$No of \$\beta^{+}\$ emitted \$= 2 \$No of \$\beta^{-}\$ emitted \$= 4\$\$\therefore\$ The resulting nucleus has atomic number z\$=z(original)-2(8)-2(1)+1(4)\$ \$\implies z=92-16 - 2 +4 = 78\$

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>>Alpha Beta and Gamma Decay
>> $Alpha Decay - Problems L1$

Alpha Decay - Problems L1

Q: The charge on alpha particles is

Alpha particles have a charge of \$+2\$,Hence, charge on an alpha particle is twice the electronic charge \$ = 2 \times 1.6 \times 10^{-19} C\$\$q = 3.2 \times 10^{-19} C\$

Q: An \$\alpha \$-particle of energy 5 MeV is scattered through \${ 180 }^{ \circ }\$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Distance of closet approx\$:-\$\${r_0} = \dfrac{{ze\,\,2e}}{{4\pi { \in _0}E}}\$\$KE=PE\$\$\dfrac{1}{2}m{v^2} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{r}\$\$r = \dfrac{{9 \times {{10}^9} \times 2 \times 92 \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}\$\$r = 5.3 \times {10^{ - 14}}\,m \approx {10^{ - 12}}\,cm\$Hence,option \$(C)\$ is correct answer.

Q: The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, \$\lambda_1/\lambda_2\$ of the photons emitted in this process is:

\$\dfrac{1}{\lambda} = R \left(\dfrac{1}{n^2_f} - \dfrac{1}{n^2_i}\right)\$here for wavelength\$\lambda_1 , n_f =3, n_i =2\$\$\dfrac{1}{\lambda_1} = R\left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right)\$\$\dfrac{1}{\lambda_1} = R\left(\dfrac{7}{9 \times 16}\right)\$here for wavelength\$\lambda_1 , n_f =2, n_i =3\$\$\dfrac{1}{\lambda_2} = R\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)\$\$= R\left(\dfrac{5}{4 \times 9}\right)\$\$\dfrac{\lambda_1}{\lambda_2} = \dfrac{\dfrac{5}{36}}{\dfrac{7}{9 \times 16}}\$Hence, from eqs. 1 and eqs. 2, we get\$= \dfrac{20}{7}\$

Q: \$\alpha \$ and \$\beta \$ - particles emitted when uranium nucleus \$_{92}U^{238}\$ decay to \$_{82}Pb^{214}\$ are

When an alpha particle is emitted from a radioactive substance its atomic number(\$Z\$) decreases by \$2\$ and mass number(\$A\$) decreases by \$4\$. When \$\beta\$ particle is emitted from a radioactive substance its atomic number(Z) increases by \$1\$ and mass number(\$A\$) does not change.Here, Uranium decays to lead its mass no changes to by \$24\$ i.e. \$\dfrac{24}{4} = 6 \alpha\$ particles are emitted and hence, its atomic number changes by 12 but the atomic numbre of \$P_b\$ is 10 this is due to emission of two \$\beta\$ particle. \$_{92}^{238}\textrm{U}\rightarrow 6\alpha_{2}^{4} + 2\beta_{1}^{0} + _{82}^{214}\textrm{Pb}\$

Q: \$10^{12} \alpha\$-particles (nuclei of helium) fall per second on a neutral sphere; calculate the time in which the sphere gets charged by \$2\mu C\$.

Total charge required \$Q= 2\mu C = 2\times 10^{-6} \ C\$Charge of one \$\alpha\$ particle \$q = 2e= 2\times 1.6\times 10^{-19} \ C\$Charge accumulated per second \$Q' = 10^{12}\times 2\times 1.6\times 10^{-19} \ C\$So, time taken \$t = \dfrac{Q}{Q'} = \dfrac{2\times 10^{-6}}{10^{12}\times 2\times 1.6\times 10^{-19}} = 6.25 \ s\$

Physics

definition

Change in nucleus due to alpha decay

When a nucleus undergoes alpha-decay, it transforms to a different nucleus by emitting an alpha-particle (a helium nucleus,

_{2}^{4} H e

).
For example, when

_{92}^{238} U

undergoes alpha-decay, it transforms to

_{90}^{234} T

_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} H e

In this process, it is observed that since

_{2}^{4} H e

contains two protons and two neutrons, the mass number and the atomic number of the daughter nucleus decreases by four and two, respectively. Thus, the transformation of a nucleus

_{Z}^{A} X

into a nucleus

_{Z - 2}^{A - 4} Y

can be expressed as

_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} H e

where

_{Z}^{A} X

is the parent nucleus and

_{Z - 2}^{A - 4} Y

is the daughter nucleus.

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A nucleus with $Z = 92$ emits the following in a sequence: $α, α, β^{-}, β^{-}, α, α, α, α; β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . The $Z$ of the resulting nucleus is

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Change in nucleus due to alpha decay

REVISE WITH CONCEPTS

Alpha Decay

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Gamma Decay

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Problem based on multiple decay

Problem on Simultaneous Radioactive Disintegration.

The charge on alpha particles is

An α-particle of energy 5 MeV is scattered through 180∘ by a fixed uranium nucleus. The distance of the closest approach is of the order of

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, λ1​/λ2​ of the photons emitted in this process is:

A nucleus with mass number 220 initially at rest emits an α particle. If the Q value of the reaction is 5.5MeV, calculate the kinetic energy of the α particle

A nucleus nm​X emits one α particle and two β particles. The resulting nucleus is

α and β - particles emitted when uranium nucleus 92​U238 decay to 82​Pb214 are

1012α-particles (nuclei of helium) fall per second on a neutral sphere; calculate the time in which the sphere gets charged by 2μC.

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

A nucleus with Z=92 emits the following in a sequence: α,α,β−,β−,α,α,α,α;β−,β−,α,β+,β+,α. The Z of the resulting nucleus is

An $α$ -particle of energy 5 MeV is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $λ_{1} / λ_{2}$ of the photons emitted in this process is:

A nucleus with mass number $220$ initially at rest emits an $α p a r t i c l e$ . If the $Q$ value of the reaction is $5.5 M e V$ , calculate the kinetic energy of the $α p a r t i c l e$

A nucleus $_{n}^{m} X$ emits one $α$ particle and two $β$ particles. The resulting nucleus is

$α$ and $β$ - particles emitted when uranium nucleus $_{92} U^{238}$ decay to $_{82} P b^{214}$ are

$1 0^{12} α$ -particles (nuclei of helium) fall per second on a neutral sphere; calculate the time in which the sphere gets charged by $2 μ C$ .

A radioactive nucleus (initial mass number A and atomic number Z) emits $3$ $α$ -particles and $2$ positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

A nucleus with $Z = 92$ emits the following in a sequence: $α, α, β^{-}, β^{-}, α, α, α, α; β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . The $Z$ of the resulting nucleus is