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marysya [2.9K]
3 years ago
6

A radioactive nuclide of atomic number emits an electron, then the daughter nuclide emits a gamma ray. What is the atomic number

of the resulting nuclide after both processes?a. Z - 1b. Z + 1c. Z + 2d. Z - 2
Physics
1 answer:
Galina-37 [17]3 years ago
3 0

Answer:

correct answer is b    Z --->Z+1

Explanation:

In the processes of radioactive decay there are three basic processes the emission of alpha particles and the emission of beta rays and the emission of ayos range

The emission of a beta ray implies the transformation of a neutral into a proton, which implies the increase of the atomic number in a unit

          Z   ---->  Z +1

the atomic mass does not change since the mass of the two particles is practically the same, to balance the reaction antineutrino must also be emitted

The daughter particle is in an execrated state and passes to its base state with the emission of a gamma ray that does not change its atomic number or its atomic mass.

Consequently, from the above the correct answer is b

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3 years ago
A coil consists of 200 turns of wire. Each turn is a square of side d = 18 cm, and a uniform magnetic field directed perpendicul
vampirchik [111]

Answer:

Induced emf in the coil, \epsilon=4.05\ volts

Explanation:

Given that.

Number of turns in the coil, N = 200

Side of square, d = 18 cm = 0.18 m

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To find,

The magnitude of the induced emf in the coil while the field is changing.

Solution,

We know that due to change in the magnetic field, an emf gets induced in the coil. The formula of induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(NBA)}{dt}

A is the ares of square

\epsilon=AN\dfrac{d(B)}{dt}

\epsilon=AN\dfrac{B_f-B_i}{t}

\epsilon=(0.18)^2\times 200 \times \dfrac{0.5-0}{0.8}

\epsilon=4.05\ volts

So, the induced emf in the coil is 4.05 volts. Hence, this is the required solution.

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3 years ago
What happens when a magenta light is shone on a green surface?
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Answer:

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