Each question contains STATEMENTS-1 (Assertion) and STATEMENT -2 (Reason). Examine the statements carefully and mark the correct answer according to the instructions given below: STATEMENT-1: Nucleus does not contain fee electrons, yet it emit beta-particles STATEMENT-2: At high n/p ratio, one neutron supposed to give 1 proton and 1 `e^(-)(beta)`.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Each question contains STATEMENTS-1 (Assertion) and STATEMENT -2 (Reas

1.	Each question contains STATEMENTS-1 (Assertion) and STATEMENT -2 (Reason). Examine the statements carefully and mark the correct answer according to the instructions given below: STATEMENT-1: Nucleus does not contain fee electrons, yet it emit beta-particles STATEMENT-2: At high n/p ratio, one neutron supposed to give 1 proton and 1 `e^(-)(beta)`.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - A

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The radiation from naturally occuring radioactive substance as seen after deflection by a magnetic field in one direction are :A. `alpha` -raysB. `beta`-raysC. both `alpha` and `beta` raysD. either `alpha` or `beta`-rays
Arrhenius studies the effect of temperature on the rate of a reaction and postulted that rate constant varies with temperature exponentially as `k=Ae^(E_(a)//RT)` . Thuis method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied. If the rate of reaction doubles for `10^(@)C` rise of temperature form 290K to 300K, the activation energy of the reaction will be approximately :A. 40 Kcal `mol^(-1)`B. 12 Kcal `mol^(-1)`C. 60 Kcal `mol^(-1)`D. 70 Kcal `mol^(-1)`
Arrhenius studies the effect of temperature on the rate of a reaction and postulted that rate constant varies with temperature exponentially as `k=Ae^(E_(a)//RT)` . Thuis method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied. The pre-exponetial factor in the Arrhenius equation of a first order reaction has the unit :A. mol `L^(-1)s^(-1)`B. L `mol^(1)s^(-1)`C. `s^(-1)`D. `dimensionaless
Arrhenius studies the effect of temperature on the rate of a reaction and postulted that rate constant varies with temperature exponentially as `k=Ae^(E_(a)//RT)` . Thuis method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied. If x is the fraction of molecules having energy greater than `E_(a)` it will be given by :A. `x=-(E_(a))/(RT)`B. In `x=-(E_(a))/(RT)`C. `x=e^(E_(a)//RT)`D. Any of these
Emission of `beta` -particle is equivatent to:A. increase of one proton onlyB. decrease of one neutron onlyC. both (a) and (b)D. none of these
Size of nucleus was obtained by the equation `r=R_(0)A^(1//3)` , Where r is the radius of nucleus of mass no A. and `R_(0)` is a constant whose valie is equal to `1.5xx10^(-15)` metre. (Given : 1 amu = `1.66xx10^(-24)g`) What is the density of a nucleus of mass number A ?A. `(4)/(3)pi(1.5xx10^(-15))^(3)` AB. `1.17xx10^(17)kg//cm^(3)`C. `1.17xx10^(-17)kg//m^(3)`D. none of these
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as `-(dN)/(dt)=lambdaN` Where `lambda=` decay constant, N= number of nuclei at time t, `N_(0)` =intial no. of nuclei. The above equation after integration can be represented as `lambda=(2.303)/(t)log((N_(0))/(N))` Calculate the half-life period of a radioactive element which remains only `1//16` of its original amount in `4740` years:A. 1185 yearsB. 2370 yearsC. `52.5` yearsD. none of these
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