Nuclear Physics

Q: Two radioactive materials A and B have decay constant 10 λ₀ and λ₀ . If initially they have same number of nuclei , find the time after which the ratio of number their undecayed nuclei will be (1/e)

Solution : Since $\large N = N_0 e^{-\lambda t }$

$\large \frac{N_A}{N_B} = \frac{N_0 e^{-10 \lambda_0 t}}{N_0 e^{-\lambda_0 t}}$

$\large \frac{N_A}{N_B} = e^{-9\lambda_0 t}$

$\large \frac{1}{e} = e^{-9\lambda_0 t}$

$\large 1 = 9 \lambda_0 t$

$\large t = \frac{1}{9 \lambda_0 } $

Q: The mean life of a radioactive substance are 1620 and 405 year for α – emission and β-emission respectively . Find out the time during which three fourth of a sample will decay if it is decaying both by α – emission and β-emission simultaneously .

Solution : When a substance decays by α – emission and β-emission simultaneously , the equivalent rate of disintegration is given by λ_eq = λ_α + λ_β

Mean life is given by $\large T_{eq} = \frac{1}{\lambda_{eq}} $

$\large T_{eq} = \frac{1}{\lambda_{eq}} $

λ_eq = λ_α + λ_β

$\large \frac{1}{T_{eq}} = \frac{1}{T_{\alpha}} + \frac{1}{T_{\beta}} $

$\large \frac{1}{T_{eq}} = \frac{1}{1620} + \frac{1}{405} $

$\large \lambda_{eq} = \frac{1}{1620} + \frac{1}{405} $

$\large \lambda_{eq} = \frac{5}{1620} year^{-1} $

As remaining sample is N₀/4

$\large N = N_0 (\frac{1}{2})^n$

$\large \frac{N_0}{4} = N_0 (\frac{1}{2})^n$

n = 2 = Number of half lives

$\large n = \frac{t}{T_{1/2}}$

$\large t = n T_{1/2}$

$\large t = n \frac{0.693}{\lambda_{eq}}$

$\large t = 2 \times \frac{0.693}{\frac{5}{1620}}$

t = 449 year

Q: The wavelength of K_α X-ray produced by an x-ray tube is 0.76Aº . The atomic number of the anti-cathode material is

(a) 82

(b) 41

(c) 20

(d) 10

Sol: For K_a X-ray line

$\large \frac{1}{\lambda_\alpha} = R(Z-1)^2 (\frac{1}{1^2} -\frac{1}{2^2})$

$\large \frac{1}{\lambda_\alpha} = \frac{3}{4}R(Z-1)^2 $

With reference to given data and putting these values

λ_α = 0.76 × 10^-10 m

R = 1.097 × 10⁷ m^-1

We get ,

$\large \frac{1}{\lambda_\alpha} = \frac{3}{4}R(Z-1)^2 $

$\large (Z-1)^2 = \frac{4}{3 \lambda_\alpha R}$

$\large (Z-1)^2 = \frac{4}{3 \times 0.76 \times 10^{-10} \times 1.097 \times 10^7 }$

(Z-1)² = 1600

Z-1 = 40

Z = 41

Correct option is (b)

Q: Show that the frequency of K_β X-ray of a material equals to the sum of frequency of K_α and L_α X-rays of the same material.

Sol: Energy of K_α X-ray is

$\large E_{K_\alpha}= E_k – E_L$

And that of K_β X-ray is

$\large E_{k_\beta} = E_k – E_M$

$\large E_{L_\alpha} = E_L -E_M$

$\large E_{k_\beta}= E_{k_\alpha} + E_{L_\alpha} $

$\large \nu_{k_\beta} = \nu_{k_\alpha} + \nu_{L_\alpha} $

Q: An X-ray tube operated at a potential difference of 40 kV , produces heat at the rate of 720 W . Assuming 0.5% of the energy of the incident electrons is converted into X-rays . Calculate
(a) the number of electrons per second striking the target.

(b) the velocity of the incident electrons.

Sol: (a) Heat produced per second at the target is

P = 0.995 V I (since 0.5% of energy is converted into X-rays)

$\large I = \frac{P}{0.995 V} = \frac{720}{0.995 \times 40 \times 10^3}$

= 0.018 A

The number of electrons per second incident on the target

$\large n = \frac{I}{e} = \frac{0.018}{1.6 \times 10^{-19}}$

= 1.1 × 10¹⁷ electrons

(b) Energy of incident electrons

$\large \frac{1}{2}m v^2 = 1 eV $

$\large v = \sqrt{\frac{2eV}{m}}$

= 1.2× 10⁸ m/s

Q: Find the maximum frequency of the x- ray emitted by an x- ray tube operating at 30 kV .

Sol: For the maximum frequency, the total kinetic energy (eV) should be converted into an x-ray photon .
Thus $\large h \nu = e V $

$\large \nu = \frac{eV}{h}$

$\large \nu = \frac{1.6 \times 10^{-19}\times 30 \times 10^3}{6.63 \times 10^{-34}}$

= 7.2×10¹⁸ Hz

Q: An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half – life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

(A) 64

(B) 90

(C) 108

(D) 120

Solution : 64 = 2⁶

Hence the material should decay for 6 half lines for a safe laboratory.

Thus answer = 18 × 6 = 108 days

Correct option is (C)

Q: The isotope ₅ B¹² having a mass 12.014 u undergoes β – decay to ₆ C¹², C has an excited state of the nucleus ( ₆ C¹² *) at 4.041 MeV above its ground state. If ₅ B¹² decays to ₆ C¹²*, the maximum kinetic energy of the β – particle in units of MeV is (1 u = 931.5 MeV/c² , where c is the speed of light in vacuum).

Solution :

₆ C¹² *) has energy 4.041 MeV

₅ B¹² → ₆ C¹² * + _-1β⁰

Suppose β– has (KE)max. = k

Total energy released, ΔE = 931 (Δm) MeV

Mass defect, Δm = (12.014 – 12) 931.5

= 13.04 MeV

Now, ΔE = 4.041 + K

K = ΔE – 4.041 = 9 MeV

Q: I¹³¹ is an isotope of Iodine that β decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with I¹³¹ is injected into the blood of a person. The activity of the amount of I¹³¹ injected was 2.4 × 10⁵ Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person’s body, and gives an activity of 115 Bq. The total volume of blood in the person’s body, in liters is approximately (you may use e^x ≈ 1 + x for |x| << 1 and In 2 ≈ 0.7).

Solution : Let V be the total blood in ml .

After 11.5 hrs activity in 2.5 ml s 115 Bq .

$\large A = A_o e^{-\lambda t }$

$\large \frac{115}{2.5 \times 10^{-3}} = \frac{2.4 \times 10^5}{V}e^{\frac{0.7}{8}(\frac{11.5}{24})} $

$\large e^{\frac{0.7}{8}(\frac{11.5}{24})} = \frac{115 V}{2.4 \times 2.5 \times 10^2}$

$\large e^{-0.0419} = 19.167 \times 10^{-2}$

1 – 0.0419 = 19.167 × 10^-2 V

$\large V = \frac{0.95 \times 100}{19.167}$

V = 5 litre

Q: In a radioactive decay chain, ₉₀ Th ²³² nucleus decays to ₈₂Pb²¹² nucleus. Let N_α and N_β be the number of α and β – particles, respectively, emitted in this decay process. Which of the following statements is (are) true ?

(A) N_α = 5

(B) N_α = 6

(C) N_β = 2

(D) N_β = 4

Solution : ₉₀ Th ²³² is converting into ₈₂Pb²¹²

Change in mass number (A) = 20 is only brought by α

no. of α particle = 20/4 = 5

Due to 5 α particle, z will change by 10 unit

Since given change is 8, therefore no. of β particle is 2

Correct options are (A) & (C)