Radioactivity is a phenomenon discovered by Henry Becquerel. The discovery of radioactivity played a significant role in the development of Nuclear Physics. While conducting an experiment connected with X-rays, Becquerel accidentally discovered that some uranium salt crystals emit certain invisible and highly penetrating radiations, which affect the photographic plates.

Radioactivity is exhibited by the nuclei of an atom as a result of nuclear instability. It is a process by which the nucleus of an unstable atom loses energy by emitting radiation. The substances exhibiting the phenomenon are called radioactive substances and the radiations emitted are called radioactive radiations. **Radioactivity is a nuclear phenomenon and it is unaffected by external factors like temperature, pressure, electric and magnetic fields**. Radioactivity can be seen in such forms

- Gamma Decay (Photons having high energy are emitted)
- Beta Decay (Emission consists of Electrons)
- Alpha Decay (Emission consists of Helium nucleus)

Download Radioactivity Previous Year Solved Questions PDF

## JEE Main Previous Year Solved Questions on Radioactivity

**Q1: A sample of radioactive material A, that has an activity of 10 mCi (1 Ci = 3.7 × 10 ^{10 }decays/s) has twice the number of nuclei as another sample of a different radioactive material B, which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively **

(a) 20 days and 5 days

(b)10 days and 40 days

(c) 20 days and 10 days

(d) 5 days and 10 days

**Solution**

R_{A} = 10 mCi, R_{B} = 20 mCi, N_{A} = 2N_{B}

R_{A}/R_{B} = λ_{A}N_{A}/λ_{B}N_{B} = [(T_{1/2})_{B}/ (T_{1/2})_{A}] x [N_{A}/N_{B}]

(½) = [(T_{1/2})_{B}/ (T_{1/2})_{A}] x 2 ⇒ (T_{1/2})_{A}= 4(T_{1/2})_{B}

**Answer: (a) 20 days and 5 days **

**Q2: Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to **

(a) 200

(b) 360

(c) 150

(d) 400

**Solution**

According to the law of radioactivity, the count rate at t = 8 seconds is

N_{1} = N_{0}e^{-λt}

dN/dt = λN =λN_{0}e^{-λt}

1600 = λN_{0}e^{0 }= λN_{0 }⇒ 100 = λN_{0}e^{-8λ }= 1600 e^{-8λ}

e^{8λ}= 16 = 24 ⇒ e^{2λ}= 2

At t = 6 sec

(dN/dt) = λN_{0}e^{-6λ }= 1600 x (e^{–2λ})^{3} = 1600 x(⅛) = 200

**Answer: (a) 200 **

**Q3: Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation? **

**Solution**

E = 13.6 (1/1 – ¼) = 13.6 x (¾) = 10.2 eV

Let us check the transitions possible on He

n = 1 or n = 2

E_{1} = 4 x 13.6 (1 – ¼) = 40.8 eV [E_{1} ＞ E, hence not possible]

n = 1 or n = 3

E_{2} = 4 x 13.6 (1 – (1/9)) = 48.3 eV [E_{2} ＞ E, hence not possible]

n = 2 or n = 3

E_{3} = 4 x 13.6 ((¼) – (1/9)) = 7.56 eV [E_{2} ＜ E, hence it is possible]

n = 2 or n = 4

E_{4} = 4 x 13.6 ((¼) – (1/6)) = 10.2 eV [E_{2} = E, hence it is possible]

Hence E_{3} and E_{4 }can be possible

**Answer: E _{3} and E_{4 }is possible**

**Q4: Two radioactive materials A and B have decay constants 10λ and λ, respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time **

(a) 1/9λ

(b) 11/10λ

(c) 1/10λ

(d) 1/11λ

**Solution**

N = N_{0}e^{-λt}

So, N_{1} = N_{0}e^{-10λt }and N_{2} = N_{0}e^{-λt }

⇒ (1/e) = (N_{1}/N_{2}) = (N_{0}e^{-10λt })/(N_{0}e^{-λt })

⇒ (1/e) = e^{-9λt }= e^{-1} = e^{-9λt }

⇒ 1 = 9λt ⇒ t = 1/9λ

**Answer: (a) 1/9λ **

**Q5: Half-lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has an equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be **

(a) 3: 8

(b) 1: 8

(c) 9: 8

(d) 8: 1

**Solution**

By the law of radioactivity N = N_{0} e^{-λt}

For nuclei A,

N_{A} = N_{0A}e^{-λt}

Or (N_{A}/N_{0A}) = (½)^{n} = (½)^{t/10} = (½)^{6 }——-(1)

N_{A}= N_{0A}/ 2^{6}

For nuclei B,

(N_{A}/N_{0A}) = (½)^{n} = (½)^{t/20} = (½)^{3}——-(2)

⇒ N_{B} = (N_{0B})/2^{3}

Ratio of nuclei decayed will be

(N’_{A}/N’_{B})= (N_{0A} – N_{A})/(N_{0B} – N_{B}) = (N_{0A}/N_{0B})[1 – (½)^{6}/1 – (½)^{3}] = 9/8

**Answer: (c) 9: 8 **

**Q6: Two radioactive substances A and B have decay constant 5λ and λ respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become (1/e) ^{2} will be **

(a) 2/λ

(b) 1/λ

(c) 1/4λ

(d) 1/2λ

**Solution**

The number of undecayed nuclei at any time t,

N = N_{0} e^{-λt}

As N_{0A} = N_{0B} (given)

So, for nuclei A and B

(N_{A}/N_{B}) = e^{(-λA + λB )t}

t = [1/(λ_{B} – λ_{A})]In(N_{A}/N_{B}) = 1/(λ – 5λ)In(1/e^{2}) = 1/2λ

**Answer: (d) 1/2λ**

**Q7: The radiation corresponding to 3➝2 transitions of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10 ^{–4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to **

(a) 1.6 eV

(b) 1.8 eV

(c) 1.1 eV

(d) 0.8 eV

**Solution**

Radius of a charged particle moving in a constant magnetic field is given by

R = (mv/qB) or R^{2} = m^{2}v^{2}/q^{2}B^{2}

R = [2m((½)mv^{2})]/q^{2}B^{2}

R = 2m(K.E)/q^{2}B^{2}

^{⇒ }K.E = (q^{2}B^{2}R^{2})/2m ^{⇒ }K.E_{max} = (q^{2}B^{2}R^{2}_{max} )/2m = 0.80 eV

Energy of photon corresponding transition from orbit 3**➝** 2 in hydrogen atom.

E = 13.6(1/2^{2}) – (1/3^{2})) = 1.89 eV

Using Einstein photoelectric equation.

E = K.E.max + Φ

⇒ 1.89 = 0.8 + Φ

⇒Φ = 1.09 ≈ 1.1 eV

**Answer (c) 1.1 eV **

**Q8: Assume that a neutron breaks into a proton and an electron. The energy released during this process is **

**(Mass of neutron = 1.6725 × 10 ^{–27} kg **

**Mass of proton = 1.6725 × 10 ^{–27} kg **

**Mass of electron = 9 × 10 ^{–31} kg) **

(a) 7.10 MeV

(b) 6.30 MeV

(c) 5.4 MeV

(d) 0.73 MeV

**Solution**

Mass defect, ∆m = m_{p} + m_{e} – m_{n}

∆m = (1.6725 x 10^{-27}) + ( 9 x 10^{-31}) – (1.6725 x 10^{-27}) Kg

∆m = 9 x 10^{-31} kg

Energy released = ∆mc^{2}

Energy released = (9 x 10^{-31}) x (3 x 10^{8})^{2} J

Energy released = [(9 x 10^{-31}) x (9 x 1016)]/[1.6 x 10^{-13 }] Mev = 0.51 MeV

**Answer: None of the given options are correct**

**Q9: The half-life period of a radioactive element X is the same as the mean lifetime of another radioactive element Y. Initially, they have the same number of atoms. Then **

(a) X and Y decay at the same rate always

(b) X will decay faster than Y

(c) Y will decay faster than X

(d) X and Y have the same decay rate initially

**Solution**

T_{1/2}, half life of X = T_{mean} ,mean life of y

Or 0.693/λ_{x} = 1/λ_{y}

λ_{x} = 0.693λ_{y}

λ_{x} ＜ λ_{y}

Rate of decay = λN

Initially, number of atoms (N) of both are equal but since λ_{x} ＜ λ_{y}, therefore y will decay at a faster rate than x

**Answer: (c) Y will decay faster than X **

**Q10: If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li ^{++} is **

(a) 30.6 eV

(b) 13.6 eV

(c) 3.4 eV

(d) 122.4 eV

**Solution**

E_{2} = (-Z^{2}E_{0})/n^{2}

E_{2} = (-(3)^{2} x 13.6)/(2)^{2}

= – 30.6 eV

Energy required = 30.6 eV

**Answer: (a) 30.6 eV **