Question

1.A radioactive sample has a half-life of 2.5 min. What fraction of the sample is left after 40 min?

2. The energy of a 2.00 keV electron is known to within

Answer 1

1. In every 2.5 minutes it becomes half. 40/2.5 =16 . Therefore after 40 minutes only 1/(2^16) th fraction will be left.

Mathematically, A = A₀(1/2)^(t/h) . Where A is the final amount, A₀ is the initial amount, t is the total time and h is the half life.

=> A = A₀(1/2)^(40/2.5)

=> A= A₀(1/2)^16

2. No matter how many times i do i am getting the answer as 2.91 * 10^10 m. Just check out with your question whether the answer is 10^-30 m or 10^-10 m.

3. Energy(E) = plancks constant(h) * frequency (f)

Considering all the units in SI system

=> 5.68 * 10^-19 = (6.63 * 10^-34) * f

=> f = (5.68/6.63) * 10^15

=> f = 0.86*10^15

Velocity of light(v) = Wavelength ( w) * frequency (f)

=> 3 * 10^8 = w * 0.86*10^15

=> w = (3/0.86) * (10 ^-7)

w = 3.49 * 10^-7 m = 349 nm which approximately equal to 350 nm

Ans(A)

4.

Mathematically, A = A₀(1/2)^(t/h) . Where A is the final amount, A₀ is the initial amount, t is the total time and h is the half life.

=> A = A₀(1/2)^(15.2/9.2)

=> A= 4*(10^18)*(1/2)^(15.2/9.2)

=> A = 1.27*10^18 atoms which is approximately equal to 1.3 * 10^18

Ans(D)

5. When a beta particle is emitted the atomic number increases by 1 and mass number doesn't change as beta is e⁰_-1. Therefore the daughter number will have 9 protons and 17 nucleons. Ans(A)

6. The force of attraction is inversely proportional to r^2, where r is the distance between the two particles. So if the force of attraction decreases by a factor of 100, r should increase by a factor of 10. Ans(D)