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: 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\$

Q: 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).

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\$

Q: 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}\$

Q: 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\$

Question 1

The charge on alpha particles is

Accepted Answer

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$

Question 2

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

Accepted Answer

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.

Question 3

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:

Accepted Answer

$\dfrac{1}{\lambda} = R \left(\dfrac{1}{n^2_f} - \dfrac{1}{n^2_i}\right)$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}$

Question 4

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 $\alpha\ particle$

Accepted Answer

$ 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$

Question 5

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

Accepted Answer

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$

Question 6

Calculate the energy released if $U^{238}$ -nucleus emits an $α -$ particle.
OR
Calculate the energy released in MeV in the following nuclear reaction
$_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} H e + Q$
Given Atomic mass of $^{238} U = 238.05079 u$
Atomic mass of $^{234} T h = 234.04363 u$
Atomic mass of alpha particle $= 4.00260 u$
$1 u = 931.5 M e V / c^{2}$
Is the decay spontaneous ? Give reasons.

Accepted Answer

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).

Question 7

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

Accepted Answer

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. $_{92}^{238}\textrm{U}\rightarrow 6\alpha_{2}^{4} + 2\beta_{1}^{0} + _{82}^{214}\textrm{Pb}$

Question 8

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

Accepted Answer

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    $t = \dfrac{Q}{Q'} = \dfrac{2\times 10^{-6}}{10^{12}\times 2\times 1.6\times 10^{-19}} = 6.25 \ s$

Question 9

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

Accepted Answer

When a radioactive nucleus emits 1 &#945;-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 $1 &#946;$-particle (positron), the atomic mass remains unchanged but the atomic number decreases by 1.So after emission of 2 positrons    $A&#34; = A' = A-12$    $Z&#34; = Z'-2&#215;1 = Z-8$So, the ratio of neutrons to protons-$\dfrac{n}{p} = \dfrac{A-Z-4}{Z-8}$

Question 10

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

Accepted Answer

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$=z(original)-2(8)-2(1)+1(4)$ $\implies  z=92-16 - 2 +4 = 78$

Alpha, Beta, and Gamma Decay

Language

Rate

REVISE WITH CONCEPTS

Plot Number of Neutrons vs Protons for Radioactive Decay

Alpha, Beta, Gamma Decay

Modes of Decay

Advanced Knowledge on Nuclear Reaction

QUICK SUMMARY WITH STORIES

Balanced Equations

Nuclear Reactions