30 + Interview Questions in MODERN PHYSICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	When a nucleus with atomic number Z and mass number A undergoes a radioactive decay process, (a) both Z and A will decrease, if the process is α decay (b) Z will decrease but A will not change, if the process is β+ decay (c) Z will increase but A will not change, if the process is β- decay (d) Z and A will remain unchanged, if the process is γ decay
Answer» Correct Answer is: (a, b, c, & d)

Discussion

2.	Let ν1 be the frequency of the series limit of the Lyman series, ν2 be the frequency of the first line of the Lyman series, and ν3 be the frequency of the series limit of the Balmer series. (a) ν1 - ν2 = ν3 (b) ν2 - ν1 = ν3 (c) ν3 = 1/2 (ν1 + ν2) (d) ν1 + ν2 = ν3
Answer» Correct Answer is: (a) ν₁ - ν₂ = ν₃ Series limit means the shortest possible wavelength (maximum photon energy), and first line means the longest possible wavelength (minimum photon energy) in the series. ν = c [1/n² - 1/m²], where c = constant. For series limit of the Lyman series, n = 1, m = ∞, ν₁ = c. For first line of the Lyman series, n = 1, m = 2, ν₂ = 3c/4 . For series limit of the Balmer series, n = 2, m ∞, ν₃ = c/4. ∴ ν₁ - ν₂ = ν₃.

2.

Let ν1 be the frequency of the series limit of the Lyman series, ν2 be the frequency of the first line of the Lyman series, and ν3 be the frequency of the series limit of the Balmer series. (a) ν1 - ν2 = ν3 (b) ν2 - ν1 = ν3 (c) ν3 = 1/2 (ν1 + ν2) (d) ν1 + ν2 = ν3

Answer»

Correct Answer is: (a) ν₁ - ν₂ = ν₃

Series limit means the shortest possible wavelength (maximum photon energy), and first line means the longest possible wavelength (minimum photon energy) in the series.

ν = c [1/n² - 1/m²], where c = constant.

For series limit of the Lyman series, n = 1, m = ∞, ν₁ = c.

For first line of the Lyman series, n = 1, m = 2, ν₂ = 3c/4 .

For series limit of the Balmer series, n = 2, m ∞, ν₃ = c/4.

∴ ν₁ - ν₂ = ν₃.

Discussion

3.	When a hydrogen atom emits a photon of energy 12.1 eV, its orbital angular momentum changes by (a) 1.05 x 10-34 J s (b) 2.11 x 10-34 J s (c) 3.16 x 10-34 J s (d) 4.22 x 10-34 J s
Answer» Correct Answer is:(b) 2.11 x 10^-34 J s Emission of a photon of 12.1 eV requires a transition from n = 3 to n = 1. Change in orbital angular momentum = h/2π (3-1) = 4/π.

3.

When a hydrogen atom emits a photon of energy 12.1 eV, its orbital angular momentum changes by (a) 1.05 x 10-34 J s (b) 2.11 x 10-34 J s (c) 3.16 x 10-34 J s (d) 4.22 x 10-34 J s

Answer»

Correct Answer is:(b) 2.11 x 10^-34 J s

Emission of a photon of 12.1 eV requires a transition from n = 3 to n = 1.

Change in orbital angular momentum = h/2π (3-1) = 4/π.

Discussion

4.	A photon of energy 10.2 eV corresponds to light of wavelength λ0. Due to an electron transition from n = 2 to n = 1 in a hydrogen atom, light of wavelength λ is emitted. If we take into account the recoil of the atom when the photon is emitted,(a) λ > λ0 (b) λ > λ0 (c) λ > λ0 (d) the data is not sufficient to reach a conclusion
Answer» Correct Answer is: (c) λ > λ₀ The total energy available from the transition = 10.2 eV = energy of emitted photon + kinetic energy of recoiling atom. ∴ energy of emitted photon < 10.2 eV. ∴ λ > λ₀.

4.

A photon of energy 10.2 eV corresponds to light of wavelength λ0. Due to an electron transition from n = 2 to n = 1 in a hydrogen atom, light of wavelength λ is emitted. If we take into account the recoil of the atom when the photon is emitted,(a) λ > λ0 (b) λ > λ0 (c) λ > λ0 (d) the data is not sufficient to reach a conclusion

Answer»

Correct Answer is: (c) λ > λ₀

The total energy available from the transition = 10.2 eV = energy of emitted photon + kinetic energy of recoiling atom.

∴ energy of emitted photon < 10.2 eV.

∴ λ > λ₀.

Discussion

5.	Which of the following pairs constitute very similar radiations? (a) Hard ultraviolet rays and soft X-rays (b) Soft ultraviolet rays and hard X-rays (c) Very hard X-rays and low-frequency γ-rays (d) Soft X-rays and γ-rays
Answer» Correct Answer is: (a) Hard ultraviolet rays and soft X-rays, (c) Very hard X-rays and low-frequency γ-rays

Discussion

6.	An orbital electron in the ground state of hydrogen has an angular momentum L1, and an orbital electron in the first orbit in the ground state of lithium has an angular momentum L2. (a) L1 = L2 (b) L1 = 3L2 (c) L2 = 3L1 (d) L2 = 9L1
Answer» Correct Answer is: (a) L₁ = L₂

Discussion

7.	A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will depend (a) only on the number of times radiation is drawn from it (b) only on the intensity of doses drawn from it (c) on both (a) and (b) (d) neither on (a) nor on (b)
Answer» Correct Answer is: (d) neither on (a) nor on (b)

Discussion

8.	Which of the following assertions are correct? (a) A neutron can decay to a proton only inside a nucleus. (b) A proton can change to a neutron only inside a nucleus. (c) An isolated neutron can change into a proton. (d) An isolated proton can change into a neutron.
Answer» Correct Answer is: (b, c)

Discussion

9.	Two identical nuclei A and B of the same radioactive element undergo β decay. A emits a β-particle and changes to A'. B emits a β-particle and then a γ-ray photon immediately afterwards, and changes to B'. (a) A' and B' have the same atomic number and mass number. (b) A' and B' have the same atomic number but different mass numbers (c) A' and B' have different atomic numbers but the same mass number. (d) A' and B' are isotopes.
Answer» Correct Answer is: (a) A' and B' have the same atomic number and mass number.

Discussion

10.	A and B are isotopes. B and C are isobars. All three are radioactive. (a) A, B and C must belong to the same element. (b) A, B and C may belong to the same element. (c) It is possible that A will change to B through a radioactive decay process. (d) It is possible that B will change to C through a radioactive decay process.
Answer» Correct Answer is: (d)

Discussion

11.	The count rate from 100 cm3 of a radioactive liquid is c. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be c/10 after three half-lives. The volume of the remaining liquid, in cm3, is (a) 20 (b) 40 (c) 60 (d) 80
Answer» Correct Answer is: (d) 80 Initial count rate (CR) for 1 cm³ of liquid = c/100. After 3 half-lives, CR for 1 cm³ of liquid = 1/8 x c/100 . Let the volume of the remaining liquid = V cm³. ∴ CR of this liquid = V x c/800 = c/10 or V = 80.

11.

The count rate from 100 cm3 of a radioactive liquid is c. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be c/10 after three half-lives. The volume of the remaining liquid, in cm3, is (a) 20 (b) 40 (c) 60 (d) 80

Answer»

Correct Answer is: (d) 80

Initial count rate (CR) for 1 cm³ of liquid = c/100.

After 3 half-lives, CR for 1 cm³ of liquid = 1/8 x c/100 .

Let the volume of the remaining liquid = V cm³.

∴ CR of this liquid = V x c/800 = c/10

or V = 80.

Discussion

12.	A radioactive nuclide can decay simultaneously by two different processes which have decay constants λ1 and λ2. The effective decay constant of the nuclide is λ.(a) λ = λ1 + λ2(b) λ = 1/2 ( λ1 + λ2)(c) 1/λ = 1/λ1 + 1/λ2(d) λ = √(λ1λ2)
Answer» Correct Answer is; (a) λ = λ₁ + λ₂

Discussion

13.	90 % of the active nuclei present in a radioactive sample are found to remain undecayed after 1 day. The percentage of undecayed nuclei left after two days will be (a) 85 % (b) 81 % (c) 80 % (d) 79 %
Answer» Correct Answer is: (b) 81 % Equal fractions decay in equal times. ∴ if a fraction 0.9 remains undecayed after 1 day, a fraction (0.9)² = 0.81 will remain undecayed after 2 days.

13.

90 % of the active nuclei present in a radioactive sample are found to remain undecayed after 1 day. The percentage of undecayed nuclei left after two days will be (a) 85 % (b) 81 % (c) 80 % (d) 79 %

Answer»

Correct Answer is: (b) 81 %

Equal fractions decay in equal times.

∴ if a fraction 0.9 remains undecayed after 1 day, a fraction (0.9)² = 0.81 will remain undecayed after 2 days.

Discussion

14.	When α, β and γ radiations pass through a gas, their ionizing powers, in decreasing order, are(a) γ, α, β(b) γ, β, α(c) α, β, γ (d) β, γ, α
Answer» Correct Answer is: (c) α, β, γ

Discussion

15.	A beam of ultraviolet light of all wavelengths passes through hydrogen gas at room temperature, in the x-direction. Assume that all photons emitted due to electron transitions inside the gas emerge in the y-direction. Let A and B denote the lights emerging from the gas in the x- and y-directions respectively.(a) Some of the incident wavelengths will be absent in A. (b) Only those wavelengths will be present in B which are absent in A. (c) B will contain some visible light. (d) B will contain some infrared light.
Answer» Correct Answer is: (a, c, & d)

Discussion

16.	An electron is in an excited state in a hydrogen-like atom. It has a total energy of -3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is λ.(a) E = 6.8 eV, λ ~ 6.6 x 10-10 m(b) E = 3.4 eV, λ ~ 6.6 x 10-10 m (c) E = 3.4 eV, λ ~ 6.6 x 10-11 m (d) E = 6.8 eV, λ ~ 6.6 x 10-11 m
Answer» Correct Answer is: (b) E = 3.4 eV, λ ~ 6.6 x 10^-10 m The potential energy = -2 x kinetic energy = -2E. ∴ total energy = -2E + E = -E = -3.4 eV or E = 3.4 eV. Let p = momentum and m = mass of the electron. ∴ E = p²/2m or p = √(2mE). de Broglie wavelength , λ = h/p = h/√(2mE) ~ 6.6 x 10^-10 m.

16.

An electron is in an excited state in a hydrogen-like atom. It has a total energy of -3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is λ.(a) E = 6.8 eV, λ ~ 6.6 x 10-10 m(b) E = 3.4 eV, λ ~ 6.6 x 10-10 m (c) E = 3.4 eV, λ ~ 6.6 x 10-11 m (d) E = 6.8 eV, λ ~ 6.6 x 10-11 m

Answer»

Correct Answer is: (b) E = 3.4 eV, λ ~ 6.6 x 10^-10 m

The potential energy = -2 x kinetic energy = -2E.

∴ total energy = -2E + E = -E = -3.4 eV

or E = 3.4 eV.

Let p = momentum and m = mass of the electron.

∴ E = p²/2m

or p = √(2mE).

de Broglie wavelength , λ = h/p = h/√(2mE) ~ 6.6 x 10^-10 m.

Discussion

17.	Let the potential energy of a hydrogen atom in the ground state be zero. Then its energy in the first excited state will be (a) 10.2 eV (b) 13.6 eV (c) 23.8 eV (d) 27.2 eV
Answer» Correct Answer is: (c) 23.8 eV

17.

Let the potential energy of a hydrogen atom in the ground state be zero. Then its energy in the first excited state will be (a) 10.2 eV (b) 13.6 eV (c) 23.8 eV (d) 27.2 eV

Answer»

Correct Answer is: (c) 23.8 eV

Discussion

18.	When an electron moving at a high speed strikes a metal surface, which of the following are possible?(a) The entire energy of the electron may be converted into an X-ray photon. (b) Any fraction of the energy of the electron may be converted into an X-ray photon. (c) The entire energy of the electron may get converted to heat. (d) The electron may undergo elastic collision with the metal surface.
Answer» Correct Answer is: (a, b, & c)

Discussion

19.	If a potential difference of 20,000 volts is applied across an X-ray tube, the cut-off wavelength will be (a) 6.21 x 10-10 m (b) 6.21 x 10-11 m (c) 6.21 x 10-12 m (d) 3.1 x 10-11 m
Answer» Correct Answer is: (b) 6.21 x 10^-11 m Photon energy at cut-off wavelength = 20,000 eV. ∴ λ_c = 1242 eV nm/20 X 10³ eV = 6. 21 x 10^-11 m.

19.

If a potential difference of 20,000 volts is applied across an X-ray tube, the cut-off wavelength will be (a) 6.21 x 10-10 m (b) 6.21 x 10-11 m (c) 6.21 x 10-12 m (d) 3.1 x 10-11 m

Answer»

Correct Answer is: (b) 6.21 x 10^-11 m

Photon energy at cut-off wavelength = 20,000 eV.

∴ λ_c = 1242 eV nm/20 X 10³ eV = 6. 21 x 10^-11 m.

Discussion

20.	The minimum wavelength of X-ray that can be produced in a Coolidge tube depends on (a) the metal used as the target (b) the intensity of the electron beam striking the target (c) the current flowing through the filament (d) the potential difference between the cathode and the anode
Answer» Correct Answer is: (d) the potential difference between the cathode and the anode

Discussion

21.	When a metal of atomic number Z is used as the target in a Coolidge tube, let ν be the frequency of the Kα line. Corresponding values of Z and ν are known for a number of metals. Which of the following plots will give a straight line?(a) ν against Z (b) 1/ν against Z(c) √ν against Z(d) ν against √Z
Answer» Correct Answer is: (c) √ν against Z From Moseley’s law, √ν = a(Z - b), where a and b are constants.

21.

When a metal of atomic number Z is used as the target in a Coolidge tube, let ν be the frequency of the Kα line. Corresponding values of Z and ν are known for a number of metals. Which of the following plots will give a straight line?(a) ν against Z (b) 1/ν against Z(c) √ν against Z(d) ν against √Z

Answer»

Correct Answer is: (c) √ν against Z

From Moseley’s law, √ν = a(Z - b), where a and b are constants.

Discussion

22.	The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 > t1). Its mean life is T.(a) A1 t1 = A2 t2(b) A1 - A2/t2 - t1 = constant(c) A2 = A1 e(t1 - t2/T)(d) A2 = A1 e(t1/Tt2)
Answer» Correct Answer is: (c) A₂ = A₁ e^{(t₁ - t₂/T)} Let A₀ = initial activity. Then, A₁ = A₀ e ^-λt₁ and A₂ = A₀ e ^-λt₂. Also, λ = 1/T ∴ A₂/A₁ = e ^-λt₂/ e ^-λt₁ e(-λt₂ + λt₁) or A₂ = A₁ e ^{λ(t₁ - t₂)} = A₁ e^{(t₁ - t₂)/T.}

22.

The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 > t1). Its mean life is T.(a) A1 t1 = A2 t2(b) A1 - A2/t2 - t1 = constant(c) A2 = A1 e(t1 - t2/T)(d) A2 = A1 e(t1/Tt2)

Answer»

Correct Answer is: (c) A₂ = A₁ e^{(t₁ - t₂/T)}

Let A₀ = initial activity.

Then, A₁ = A₀ e ^-λt₁ and A₂ = A₀ e ^-λt₂. Also, λ = 1/T

∴ A₂/A₁ = e ^-λt₂/ e ^-λt₁ e(-λt₂ + λt₁)

or A₂ = A₁ e ^{λ(t₁ - t₂)} = A₁ e^{(t₁ - t₂)/T.}

Discussion

23.	The penetrating powers of α, β and γ radiations, in decreasing order, are (a) γ, α, β(b) γ, β, α (c) α, β, γ (d) β, γ, α
Answer» Correct Answer is: (b) γ, β, α

23.

The penetrating powers of α, β and γ radiations, in decreasing order, are (a) γ, α, β(b) γ, β, α (c) α, β, γ (d) β, γ, α

Answer»

Correct Answer is: (b) γ, β, α

Discussion

24.	If a beam consisting of α, β and γ radiations is passed through an electric field perpendicular to the beam, the deflections suffered by the components, in decreasing order, are(a) α, β, γ (b) α, γ, β (c) β, α, γ (d) β, γ, α
Answer» Correct Answer is: (c) β, α, γ

Discussion

25.	Let u denote one atomic mass unit. One atom of an element of mass number A has mass exactly equal to A u (a) for any value of A (b) only for A = 1 (c) only for A = 12 (d) for any value of A provided the atom is stable
Answer» Correct Answer is: (c) only for A = 12

Discussion

26.	A fraction f1 of a radioactive sample decays in one mean life, and a fraction f2 decays in one half-life. (a) f1 > f2 (b) f1 > f2(c) f1 = f2 (d) May be (a), (b) or (c) depending on the values of the mean life and half-life.
Answer» Correct Answer is: (a) f₁ < f₂ Mean life T = 1/λ and half-life T_1/2 = In 2/λ = 0.693/λ. ∴ T > T_1/2. Greater fraction will decay in longer time.

26.

A fraction f1 of a radioactive sample decays in one mean life, and a fraction f2 decays in one half-life. (a) f1 > f2 (b) f1 > f2(c) f1 = f2 (d) May be (a), (b) or (c) depending on the values of the mean life and half-life.

Answer»

Correct Answer is: (a) f₁ < f₂

Mean life T = 1/λ and half-life T_1/2 = In 2/λ = 0.693/λ.

∴ T > T_1/2. Greater fraction will decay in longer time.

Discussion

27.	In a radioactive series, 92 238U changes to 82 206Pb through n1 α-decay processes and n2 β-decay processes.(a) n1 = 8, n2 = 8 (b) n1 = 6, n2 = 6 (c) n1 = 8, n2 = 6 (d) n1 = 6, n2 = 8
Answer» Correct Answer is: (c) n₁ = 8, n₂ = 6 n₂= 238 - 206/4 = 8. n₁ = n₂ x 2 - (92 - 82) = 6.

27.

In a radioactive series, 92 238U changes to 82 206Pb through n1 α-decay processes and n2 β-decay processes.(a) n1 = 8, n2 = 8 (b) n1 = 6, n2 = 6 (c) n1 = 8, n2 = 6 (d) n1 = 6, n2 = 8

Answer»

Correct Answer is: (c) n₁ = 8, n₂ = 6

n₂= 238 - 206/4 = 8.

n₁ = n₂ x 2 - (92 - 82) = 6.

Discussion

28.	Three-fourths of the active nuclei present in a radioactive sample decay in 3/4 s. The half-life of the sample is(a) 1 s (b) 1/2 s (c) 3/4 s (d) 3/8 s
Answer» Correct Answer is: (d) 3/8 s

Discussion

29.	If the potential difference applied across a Coolidge tube is increased, (a) the wavelength of the Kα line will increase (b) the wavelength of the Kβ line will decrease(c) the difference in wavelength between the Kα and Kβ lines will decrease (d) none of the above
Answer» Correct Answer is: (d) none of the above

Discussion

30.	The diagram alongside shows three resistances of 1 Ω, 3 Ω and 4 Ω connected to a cell of e.m.f. (electromotive force) 2 V and internal resistance 2.0 Ω.a. Calculate the total resistance of the circuit.b. What reading will be shown by the ammeter?c. What reading will be shown by the ammeter if a similar cell is connected in series with the given cell?
Answer» (a) 2 ohm (b) 1 A (c) 0.5 A

Discussion

Explore topic-wise InterviewSolutions in .

Let ν1 be the frequency of the series limit of the Lyman series, ν2 be the frequency of the first line of the Lyman series, and ν3 be the frequency of the series limit of the Balmer series. (a) ν1 - ν2 = ν3 (b) ν2 - ν1 = ν3 (c) ν3 = 1/2 (ν1 + ν2) (d) ν1 + ν2 = ν3

When a hydrogen atom emits a photon of energy 12.1 eV, its orbital angular momentum changes by (a) 1.05 x 10-34 J s (b) 2.11 x 10-34 J s (c) 3.16 x 10-34 J s (d) 4.22 x 10-34 J s

Which of the following pairs constitute very similar radiations? (a) Hard ultraviolet rays and soft X-rays (b) Soft ultraviolet rays and hard X-rays (c) Very hard X-rays and low-frequency γ-rays (d) Soft X-rays and γ-rays

An orbital electron in the ground state of hydrogen has an angular momentum L1, and an orbital electron in the first orbit in the ground state of lithium has an angular momentum L2. (a) L1 = L2 (b) L1 = 3L2 (c) L2 = 3L1 (d) L2 = 9L1

Which of the following assertions are correct? (a) A neutron can decay to a proton only inside a nucleus. (b) A proton can change to a neutron only inside a nucleus. (c) An isolated neutron can change into a proton. (d) An isolated proton can change into a neutron.

The count rate from 100 cm3 of a radioactive liquid is c. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be c/10 after three half-lives. The volume of the remaining liquid, in cm3, is (a) 20 (b) 40 (c) 60 (d) 80

A radioactive nuclide can decay simultaneously by two different processes which have decay constants λ1 and λ2. The effective decay constant of the nuclide is λ.(a) λ = λ1 + λ2(b) λ = 1/2 ( λ1 + λ2)(c) 1/λ = 1/λ1 + 1/λ2(d) λ = √(λ1λ2)

90 % of the active nuclei present in a radioactive sample are found to remain undecayed after 1 day. The percentage of undecayed nuclei left after two days will be (a) 85 % (b) 81 % (c) 80 % (d) 79 %

When α, β and γ radiations pass through a gas, their ionizing powers, in decreasing order, are(a) γ, α, β(b) γ, β, α(c) α, β, γ (d) β, γ, α

Let the potential energy of a hydrogen atom in the ground state be zero. Then its energy in the first excited state will be (a) 10.2 eV (b) 13.6 eV (c) 23.8 eV (d) 27.2 eV

If a potential difference of 20,000 volts is applied across an X-ray tube, the cut-off wavelength will be (a) 6.21 x 10-10 m (b) 6.21 x 10-11 m (c) 6.21 x 10-12 m (d) 3.1 x 10-11 m

The minimum wavelength of X-ray that can be produced in a Coolidge tube depends on (a) the metal used as the target (b) the intensity of the electron beam striking the target (c) the current flowing through the filament (d) the potential difference between the cathode and the anode

The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 &gt; t1). Its mean life is T.(a) A1 t1 = A2 t2(b) A1 - A2/t2 - t1 = constant(c) A2 = A1 e(t1 - t2/T)(d) A2 = A1 e(t1/Tt2)

The penetrating powers of α, β and γ radiations, in decreasing order, are (a) γ, α, β(b) γ, β, α (c) α, β, γ (d) β, γ, α

If a beam consisting of α, β and γ radiations is passed through an electric field perpendicular to the beam, the deflections suffered by the components, in decreasing order, are(a) α, β, γ (b) α, γ, β (c) β, α, γ (d) β, γ, α

Let u denote one atomic mass unit. One atom of an element of mass number A has mass exactly equal to A u (a) for any value of A (b) only for A = 1 (c) only for A = 12 (d) for any value of A provided the atom is stable

A fraction f1 of a radioactive sample decays in one mean life, and a fraction f2 decays in one half-life. (a) f1 &gt; f2 (b) f1 &gt; f2(c) f1 = f2 (d) May be (a), (b) or (c) depending on the values of the mean life and half-life.

In a radioactive series, 92 238U changes to 82 206Pb through n1 α-decay processes and n2 β-decay processes.(a) n1 = 8, n2 = 8 (b) n1 = 6, n2 = 6 (c) n1 = 8, n2 = 6 (d) n1 = 6, n2 = 8

Three-fourths of the active nuclei present in a radioactive sample decay in 3/4 s. The half-life of the sample is(a) 1 s (b) 1/2 s (c) 3/4 s (d) 3/8 s

If the potential difference applied across a Coolidge tube is increased, (a) the wavelength of the Kα line will increase (b) the wavelength of the Kβ line will decrease(c) the difference in wavelength between the Kα and Kβ lines will decrease (d) none of the above

The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 > t1). Its mean life is T.(a) A1 t1 = A2 t2(b) A1 - A2/t2 - t1 = constant(c) A2 = A1 e(t1 - t2/T)(d) A2 = A1 e(t1/Tt2)

A fraction f1 of a radioactive sample decays in one mean life, and a fraction f2 decays in one half-life. (a) f1 > f2 (b) f1 > f2(c) f1 = f2 (d) May be (a), (b) or (c) depending on the values of the mean life and half-life.