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

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