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Lunna [17]
3 years ago
14

A curium-242 (Z= 96) can be produced by positive-ion bombardment when an alpha particle collides with which of the following nuc

lei? Note: A neutron is also a product of this bombardment, in addition to the curium-242.
A. Pu -239
B. U -239
C. Am-241
D. Cf-249
E. Pu-241
Physics
1 answer:
meriva3 years ago
4 0

Answer:

A. Pu -239

Explanation:

An isotope is an element with the same atomic number but different mass number. most isotope are unstable, having short half life.

Curium-242 is an isotope produced when Plutonium 239 is bombarded by an alpha particle. This reaction between Plutonium 239 and alpha particle gives curium-242, neutron and a high amount of energy as the products.

Curium oxidizes easily, and it is a dangerous metal which can cause cancer initiation when absorbed by biological materials e.g bones or tissue.

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3 years ago
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

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