A1.00 * 10^-6 -g sample of nobelium, 254/102 no, has a half-life of 55 seconds after it is formed. what is the percentage of 254/102 no remaining at the following times? a) 5.0 min after it formsb) 1.0 h after it forms

2.2 % and 0 %

Explanation:

The equation we will be using to solve this question is:

N/N₀  = e⁻λ t

where  N₀ : Number of paricles at t= 0

            N=  Number of particles after time t

             λ= Radioactive decay constant

             e= Euler´s constant

We are not given λ , but it can be determined from the half life with the equation:

λ = 0.693 / t 1/2 where t 1/2 is the half-life

Substituting our values:

λ = 0.693 / 55 s = 0.0126/s

a) For t = 5 min = 300 s

N / N₀ = e^-(0.0126/s x 300 s) = e^-3.8 = 0.022 = 2.2 %

b) For t = 1 hr = 3600 s

N / N₀ =  e^-(0.0126/s x 3600 s) = 2.9 x 10 ⁻²⁰ = 0 % (For all practical purposes)

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j