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Alika [10]
3 years ago
9

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

Physics
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X

^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

^{22}_{11}Na\Rightarrow ^{22}_{11-1}X

^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

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n=no of electron

Q=charge of Electron

A=cross-section

If area of cross-section decreases gradually then drift velocity will increase because drift velocity is inversely proportional to Area of cross-section

       

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What do work and energy have in common
spayn [35]

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Explanation:

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3 years ago
The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co
Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density = 8.4\times10^{28}\ m^3

Current i = 4.6\times10^{18}\ electron/s

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

v_{d}=\dfrac{I}{neA}

v_{d}=\dfrac{Ne}{tne\times\pi r^2}

Put the value into the formula

v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}

v_{d}=4.842\times10^{-5}\ m/s

We need to calculate the time

Using formula for time

v_{d}=\dfrac{l}{t}

t=\dfrac{l}{v_{d}}

Where, l = length

v_{d} = drift velocity

Put the value into the formula

t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}

t=6402.31\ sec

t=106.7\ minute

Hence, The time is 106.7 minute.

7 0
3 years ago
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