NucleiNEET Physics · Class 12 · NCERT Chapter 13

Solution

The energy released in the reaction is given by $Q=(B.E{.}_{{\text{Li}}_3^7}-2\cdot B.E{.}_{{\text{He}}_2^4})+B.E{.}_{{\text{H}}_1^1}$. Substituting the values, $Q=(5.60\times 7-2\times 7.06\times 4)+0=39.2-56.48+0=-17.28\text{MeV}$. Since the reaction is exothermic, $Q=17.3\text{MeV}$, so option (d) is correct.

Solution

The ratio of $X$ to $Y$ is 1:7, indicating that 8 parts of the original $X$ have decayed to 7 parts of $Y$. This corresponds to 3 half-lives, since $2^3=8$. The age of the rock is $3\times 1.4\times 10^9=4.2\times 10^9$ years, so option (c) is correct.

Solution

U→ Th+α KETh= P2 2mTh ,KEα= P2 2mα Since mα is less so KEα will be mone.

Solution

At the distance of lowest approach, total K.E. of 𝛼-particle changes to P.E. so 1 2mv2 = KQ.q r = K(ze)(2e) r r =4K ze2 mv2 ⇒ r ∝1 m www.vedantu.com 16 r ∝ 1 m

Solution

No option is correct If we take 1A B N Ne = Then 8 t A tB N e N e -λ -λ= 71 tee -λ= –1 = –7λt t = 1 7λ

Solution

The number of nuclei remaining after time $t$ is given by $N=N_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life. Substituting, $150=600{\left(\frac{1}{2}\right)}^{t/10}\Rightarrow {\left(\frac{1}{2}\right)}^{t/10}=\frac{1}{4}\Rightarrow t/10=2\Rightarrow t=20\text{minutes}$, so option (c) is correct.

Solution

(i) Req = 𝑟 3+ 𝑟 3= 2𝑟 3 (ii) R′eq = 𝑟 2+𝑟= 3𝑟 2 𝑃𝑖 𝑃𝑖𝑖 = 𝑉2/𝑅𝑒𝑞 𝑉2/𝑅′𝑒𝑞= 3𝑟/2 2𝑟/3= 9 4

Solution

The nuclear reaction is ${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{36}^{89}\text{Kr}+3{}_0^1\text{n}{+}_{56}^{144}\text{Ba}$. The mass and charge numbers must be conserved, so the remaining product is ${}_{56}^{144}\text{Ba}$, making option (a) correct.

Solution

The energy equivalent of mass is given by $E=m{c}^2$. Substituting $m=0.5\times 10^{-3}\text{ kg}$ and $c=3\times 10^8\text{ m/s}$, we get $E=0.5\times 10^{-3}\times (3\times 10^8{)}^2=1.5\times 10^{13}\text{ J}$. Therefore, option (c) is correct.

Solution

The sequence of decays involves a change in atomic number and mass number. An $\alpha$ decay reduces the atomic number by 2 and the mass number by 4, while ${\beta }^{-}$ decay increases the atomic number by 1 and ${\beta }^{+}$ decay decreases the atomic number by 1. The sequence A $\to$ B $\to$ C $\to$ D can be explained by an $\alpha$ decay followed by a ${\beta }^{-}$ decay and then a ${\beta }^{+}$ decay, so option (a) is correct.

Solution

The fraction of original activity remaining after time $t$ is given by $N=N_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$. Substituting $t=150\text{hours}$ and $T_{1/2}=100\text{hours}$, $N=N_0{\left(\frac{1}{2}\right)}^{150/100}=N_0{\left(\frac{1}{2}\right)}^{1.5}=N_0\left(\frac{1}{2\sqrt{2}}\right)=\frac{N_0}{2\sqrt{2}}=\frac{N_0}{2^{1.5}}=\frac{N_0}{2^{3/2}}$. Therefore, the fraction is $\frac{1}{2^{3/2}}$, which matches option (b).

Solution

The total binding energy before fission is $240\times 7.6\text{MeV}=1824\text{MeV}$. After fission, the total binding energy is $2\times (120\times 8.5\text{MeV})=2040\text{MeV}$. The gain in binding energy is $2040\text{MeV}-1824\text{MeV}=216\text{MeV}$, so option (d) is correct.

Solution

Tritium undergoes beta decay, emitting a beta particle (${\beta }^{-}$). This is consistent with the nuclear decay processes described in NCERT XII chapter Nuclei, so option (a) is correct.

Solution

The nuclear reaction is given as 22 0 11 1Na +→ + +νA Z Xe From conservation of atomic number 11 = Z + 1 ⇒ Z = 10 ⇒ Ne From conservation of mass number 22 = A + 0 ⇒ A = 22 ∴ 22 10Ne=A Z X

Solution

Radius of nuclei with mass number A varies as R = R0A1/3 1/ 3 1 2 5125 464 R R = = = 5 : 4 CHEMISTRY SECTION-A

Solution

The activity of a radioactive substance drops to $\left(\frac{1}{16}\right)$th of its initial value after 4 half-lives. Given the half-life is 20 minutes, $4\times 20=80$ minutes, so option (a) is correct.

Solution

286290 286 286 2868082 79 80 81 eeX Y Z P Q + − −αβ⎯⎯ → ⎯⎯⎯ → ⎯⎯⎯ → ⎯⎯⎯ → A → 286 Z = 81