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3 years ago

When aluminum-27 is bombarded with a neutron, a gamma ray is emitted. what radioactive isotope is produced?

Physics

2 answers:

alexandr402 [8]3 years ago

8 0

Answer:

Go to the excited aluminum-28 isotope

Explanation:

In the processes of nuclear reactions, fundamental changes occur at the level of the atomic nucleus.

Analyze the characteristics of the neutron that has approximately the mass of the proton and has no electric charge, so when being absorbed by the aluminum core the atomic number should not be changed, but the atomic mass should increase by one unit.

The new atom formed is in an excited state and enters the base state with the emission of a high-energy radiation, gamma ray that does not transform the atom

Tthe reaction is:

⁷A₂₇ + ⁰n₁ → ⁷Al₂₈ˣ

The symbol "ˣ" means excited

⁷Al₂₈ˣ → ⁷Al₂₈ + γ

I am Lyosha [343]3 years ago

5 0

Aluminium 28 i think thats the best isotope for this

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3 years ago

Which phrase is the best definition of energy

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Answer:

The capacity for doing work.

Explanation:

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3 years ago

What is the current that reverses direction in a regular patter called

Oxana [17]

Answer:

Current that reverses direction in the regular pattern is called an alternating current, abbreviated as 'AC'.

Explanation:

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3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are

irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

We can now find the acceleration:

F = m * a

19.62 = 3 * a

a = 6.54 m/s^2

We now use the following equation of motion to get the time taken to travel 1 meter:

$s=u*t+\frac{1}{2} (a*t^2)$

$1=0*t+\frac{1}{2} (6.54*t^2)$

t = 0.553 seconds

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3 years ago