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balandron [24]
3 years ago
5

The half-life for radioactive decay (a first-order process) is 24,000 years. How many years does it take for one mole of this ra

dioactive material to decay so that just one atom remains?
Physics
1 answer:
iren [92.7K]3 years ago
4 0

Answer: it will take 1.9 × 10⁶ yrs for one mole of the radioactive material to decay so that just one atom remains

Explanation:

Given that;

Half life t_1/2 = 24,000 years

initial amount of radio active element = 1 mole

now one mole of a substance has Avagadro number of atoms which is initially 6.023 × 10²³ which are present and finally only 1 atom is left

using the formula

t_1/2 = 0.693/k

In[ N₀/N_t] = kt

N₀ is the initial amount, N_t is the amount left after t time, t is the time, k is the rate constant.

now we substitute

t_1/2 = 0.693/k

k = 0.693/ t_1/2

k = 0.693 / 24,000 years

k = 0.000028875 yr⁻¹

so

In[ N₀/N_t] = kt

t = 1/0.000028875 yr⁻¹ In [ 6.023 × 10²³ / 1 ]

= 34632.0346 In[6.023 × 10²³]

= 34632.0346 × 54.7550

= 1896277.0545 yrs ≈ 1.9 × 10⁶ yrs

Therefore it will take 1.9 × 10⁶ yrs for one mole of the radioactive material to decay so that just one atom remains

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2 years ago
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max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

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You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

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x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

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vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

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3 years ago
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