Suppose protons were emitted rather than electrons then it affects the experiment as the mean velocity of proton will be less then mean velocity electron .
The mass of proton is greater than the mass of electrons but charge of electron is equal to the charge of proton . So , due to difference in the mass of electron and proton there will be some effects. We can conserve the electric energy which is equal to the qe .
The kinetic energy = 1/2
Now changing the electric potential energy into kinetic energy
v = √2qe/m
The mean velocity of proton will be less then mean velocity electron .
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Answer:
Option C is correct.
t = 1.95 billion years.
Explanation:
Radioactive decay follows a first order reaction kinetics.
On solving the dynamic equation (the differential equation), this is obtained
C(t) = C₀ e⁻ᵏᵗ
C(t) = amount of radioactive material remaining after time t = 37.5%
C₀ = Initial amount of radioactive material = 100%
t = time that has passed = ?
k = decay constant.
For a first order reaction, the decay constant is related to the half life through the relation
k = (In 2)/T
T = half life = 1.38 billion years
k = (In 2)/1.38
k = 0.5023 per billion years.
C(t) = C₀ e⁻ᵏᵗ
0.375 = e⁻ᵏᵗ
e⁻ᵏᵗ = 0.375
In e⁻ᵏᵗ = In 0.375 = -0.981
-kt = -0.981
t = (0.981/0.5023) = 1.95 billion years.
Hope this Helps!!!
Molecular weight of (NH4)2 CO3 = 96 gm. 96 gm (NH4)2CO3 = 1 mol of (NH4)2CO3 = 2 moles of NH4.
Therefore 1.3 gm of (NH4)2CO3 = 1.3x2/ 96 = 0.027 moles.
Answer:
lighting a match or water and salt
Ok so I’m not sure in this one but I think it’s c
