MCQ PROBLEM ON NUCLEAR PHYSICS

 

Q1. The radius of Ge nucleus is measured to be twice the radius of ⁹₄Be . How many nucleons are there in Ge nucleus?

 (a) 64                     (b) 72                     (c) 82                       (d) 86

     Ans:  R = R₀ (A)^1/3  

         R_Ge=2*R_Be             R₀(A_Ge)^1/3  = 2*R₀(9)^1/3  

          A_Ge =8*9=72 .

 

Q2. The radius of a ⁶⁴₂₉X  nucleus is measured to be  4.8 *10^-13 cm . The radius of a ²⁷₁₂Y nucleus can be estimated to be

 (a)2*10^-13 cm     (b)4*10^-13 cm      (c)6*10^-13 cm         (d)8*10^-13 cm

 Ans: Since   R = R₀ (A)^1/3  

                  R_Y/R_X =(A_y/A_X)^1/3 =(27/64)^1/3  

                    R_Y/R_X =3/4 

                     R_Y =(3/4) R_X  =3.6*10^-13 cm = 4*10^-13 cm                                  

 

 Q3. According to the empirical observations of charge radii, a ¹⁶₈X nucleus is spherical and has charge radius R and a volume V= (4/3)pR³. Then the volume of the ¹²⁸₅₄Y nucleus, is

   (a) 1.5V                                  (b) 2V                                       (c) 6.5V                                (d) 8V

    Ans: V= (4/3)pR³ =(4/3)p(R₀)³A

                =(4/3)p(R₀)³16

        V’=(4/3)p(R₀)³128=(V/16)*128=8V. 

 

 Q4. A radioactive sample containing N₀ nuclei emits N α-particle per second on decaying. The half life of the sample is 0.693 N/ N₀.

(a) 0.693 N/ N₀                                  (b) 0.693 N₀/N                           (c)1.44N/N                    (d)0 1.44 N₀/N        

Ans:  R= Nλ        ⇒N= N₀λ

           ⇒T_1/2  =0.693/λ =0.693 N₀/N        

 

Q5. According to measurements by Rutherford and Geiger, one gram of radium-226 emits in one second 3.7*10¹⁰ alpha particles. The half life of radium is

 (a) 400 years                      (b) 800 years                     (c) 1600 years                  (d) 3200 years

Ans: Number of radium atoms in one gram of radium= (1/226)*6.02*10²³ =2.7*10²¹

         Decay constant λ=-(dN/N)-1

                            λ = 3.7*10¹⁰/2.7*10²¹=1.37*10^-11 sec^-1

          Thus Half Life T_1/2=0.693/λ=0.693/1.37*10^-11 =5*10¹⁰sec

                           = 5*10¹⁰ /(365*24*60*60)=1600 years

 

Q6. A radioactive sample contains 3*10^-9 kg  of active gold ²⁰⁰Au , whose half life is 48 min . Then the activity of the radon sample is

    (a) 55 Ci                       (b) 57 Ci                           (c) 59 Ci                            (d) 61 Ci

Ans: Decay constant λ=0.693/T_1/2 =0.693/(48*60)=2.406*10^-4 sec^-1

              Number of atoms in 3*10^-9 kg   is N=(3*10^-9 kg/200)*6.023*10²³ =9.04*10¹⁵ atoms

 Hence, activity   R= Nλ=2.406*10^-4 * 9.04*10¹⁵    

                          =2.18*10¹² decay/sec

                           1.0Ci =3.7*10¹⁰deacy/sec

                        2.18*10¹² decay/sec  =  2.18*10¹²/3.7*10¹⁰=59Ci

 

 

 Q7. The radio isotope ¹⁴C maintains a fixed proportion in a living entity by exchanging carbon with the atmosphere. After it dies exchange ceases and proportion of  ¹⁴C decreases continuously as ¹⁴C beta decays with half life of 5500 years . Estimate the age of the dead tree whose present activity is 1/3 of initial activity.

  (a) 8717 years              (b) 6520 years                        (c) 5500 years            (d) 4500 years

Ans: R= R₀e^-λt  

                The age of the dead tree is t= (1/λ) ln(R₀/R)

                                = (T_1/2/0.693) ln(R₀/R)

                                 = (5500/0.693) ln(3/1)

                                  =8717 years

 Q8. A radioactive sample emits n β-particles in 2 sec . In next 2 sec it emits 0.75n β-particles, then the mean life of the sample is(ln2 = 0.693, ln3 =1.0986)

 (a)2 sec                        (b)5 sec                        (c)7 sec                           (d) 9 sec

Ans: Let N₀ be the number of initial number of nuclei. Then

           n = N₀ - N₀e^-2λ = N₀ (1- e^-2λ)

                         0.75n = N₀e^-2λ - N₀e^-2λ e^-2λ = N₀e^-2λ (1- e^-2λ)

                       0.75n/n = N₀e^-2λ (1- e^-2λ)/ N₀ (1- e^-2λ)

                        e^-2λ =3/4

                         2λ = 2 *ln2 – ln3

                             λ = 0.1438sec^-1

                                 T=1/ λ = 7sec