If a nucleus decays by successive b, a, a emissions, its mass number will Group of answer choices decrease by seven. decrease by
1 answer:
Answer:
The mass number will decrease by eight (8).
Explanation:
Given;
successive beta (b), alpha (a), alpha (a) emissions.
Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 .
Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2.
For the given question, a successive beta (b), alpha (a), and alpha (a) emissions = (0) + (-4) + (-4) = -8
Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).
<u>To learn more, please visit:</u> brainly.com/question/23276969
You might be interested in
Answer:
on increasing pressure, temperature will also increase.
Explanation:
Considering the ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.
P ∝ T
Also,
Also, using Gay-Lussac's law,
Thus, on increasing pressure, temperature will also increase.
Answer:
at r < R;
at 2R > r > R;
at r >= 2R
Explanation:
Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,
(integral over a closed surface)
where,
= Electric field
= charged enclosed within the closed surface
= permittivity of free space
Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .
at r < R,
= 0 and hence
= 0 (since the sphere is conducting, all the charges get repelled towards the surface)
at 2R > r > R,
= Q,
therefore,
(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to )
or,
at r >= 2R
= 2Q
Hence, by similar calculations, we get,