Manganese-56 is a beta emitter with a half-life of 2.6 hr. What is the mass of manganese-56 in a 49.7-mg sample of the isotope after 10.4 hr?

3.11 mg

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 2.6 hr

Original amount (N₀) = 49.7 mg

Time (t) = 10.4 hr

Amount remaining (N) =.?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.6 hr

Time (t) = 10.4 hr

Number of half-lives (n) =?

n = t / t½

n = 10.4 / 2.6

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the amount remaining. This can be obtained as illustrated below:

Original amount (N₀) = 49.7 mg

Number of half-lives (n) = 4

Amount remaining (N) =.?

N = 1/2ⁿ × N₀

N = 1/2⁴ × 49.7

N = 1/16 × 49.7

N = 0.0625 × 49.7

N = 3.11 mg

Therefore, the amount remaining is 3.11 mg

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