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3 years ago

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When gamma radiation is released by an atom,. A) it is the only radiation emitted. B) it only ever accompanies beta radiation. C

) it only ever accompanies alpha radiation. D) it usually accompanies beta or alpha radiation..

2 answers:

Sidana [21]3 years ago

7 0

Hello!

The gamma radiation haven't mass or charge. When it is released, it usually accompanies beta or alpha radiation. Letter d).

Hugs!

musickatia [10]3 years ago

6 0

Answer: D) it usually accompanies beta or alpha radiation..

Explanation:-

Gamma decay : In this process, the unstable nuclei transformed into stable nuclei by releasing gamma rays along with beta or alpha decay. It is observed when a nucleus is formed in an excited state and then decays to its ground state by emitting an electromagnetic radiation.

General representation of an element is given as: $_Z^A\textrm{X}$

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

General representation of gamma decay :

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$

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30 points, brainiest, high school physics problem, please help with explanations.

Ymorist [56]

Answer:

I think the answer is C too

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3 years ago

What are the steps that produce electricity from this fuel

deff fn [24]

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4 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0

Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

x = 0.135 cm

Given parameters

The mass m = 220 g = 0.220 kg

The spring cosntnate3 k = 7.0 N / m

Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

The position where the kinetic and potential energy are equal

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

x = A cos wt + fi

w² = $\frac{k}{m}$

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

v = $\frac{dx}{dt}$

let's evaluate

v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

K = U

½ m v² = m g x

we substitute

½ A² w² sin² wt = g A cos wt

sin² wt = $\frac{2g}{A}$ cos wt

let's calculate

w = $\sqrt{\frac{7}{0.220} }$

w = 5.64 rad / s

sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

sin² 5.64t = 376.92 cos 5.64 t

1 - cos² 5.64t = 376.92 cos 5.64t

cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

x = cos 5.64t

x²- 376.92 x - 1 = 0

x = 0.026

cos 5.64t = 0.026

Let's find the displacement for this time

x = 5.2 10-2 0.026

x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

x = 0.135 cm

Learn more here: brainly.com/question/15707891

8 0

3 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

4 years ago