Question

complete the following nuclear change
please answer fast​

Answer 1

Given that :-

⠀⠀⠀⠀

$\sf {\overline{\underline{ }}} X_1 \xrightarrow{\beta} {\overline{\underline{}}} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3$

⠀⠀⠀⠀

To Find :-

⠀⠀

Complete equation = ??

⠀⠀⠀⠀

Solution :-

⠀⠀⠀⠀

We know that atomic number is represented by Z and the mass no. is represented by A .

⠀⠀⠀⠀

Therefore ;

$\sf {}^A _ Z X_1 = {\overline{\underline{}}}X_1$ --- ( i )

⠀⠀⠀⠀

• Due to the emission of a β - particle , the atomic number get increased by 1 , but the mass number remains unchanged

⠀⠀⠀⠀

$\sf {\overline{\underline{}}}X_2 = {} ^{A+1} _ {Z} X_2$ --- ( ii )

⠀⠀⠀⠀

• Due to the alpha decay the atomic number is decreased by 2 and the mass number is reduced by 4 .

⠀⠀⠀⠀

$\sf {}^{238} _ {75} X_3= {} ^{A+1-4} _ {Z-2} X_3$ --- ( iii )

⠀⠀⠀⠀

Now the equation becomes ;

⠀⠀⠀⠀

$\sf {} ^A _ Z X_1 \xrightarrow{\beta} {}^{A+1} _{Z} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3$

⠀⠀⠀⠀

⠀⠀⠀⠀

From eq ( iii )

⠀⠀⠀⠀

Firstly ; We observe that :-

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A + 1 - 4 = 238

⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A - 3 = 238

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A = 238 + 3

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A = 241

⠀⠀⠀⠀

$\sf \: \: \: \: \: \: \:\: \: {\boxed{\large{A = 241}}}$

⠀⠀⠀⠀

⠀⠀⠀⠀

Secondly ; we observe that :-

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀Z - 2 = 75

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀Z = 75 + 2

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀Z = 77

⠀⠀⠀⠀

$\sf \: \: \:\: \: \: \: \: \: {\boxed{\large{Z = 77}}}$

⠀⠀⠀⠀

Now ; We can put the respective value in the final equation

⠀⠀⠀⠀

$\sf {} ^A _ Z X_1 \xrightarrow{\beta} {}^{A+1} _{Z} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3$

⠀⠀⠀⠀

$\sf {} ^{241} _ {77} X_1 \xrightarrow{\beta} {}^{241+1} _{77} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3$

⠀⠀⠀⠀

$\sf {} ^{241} _{ 77 }X_1 \xrightarrow{\beta} {}^{242} _{77} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3$

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⠀⠀⠀⠀

More information about the equation :-

⠀⠀⠀⠀

• X₁ and X₂ are Isotopes

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Reason :- Becoz they have same atomic number but diff. mass number

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Extra Information

⠀⠀⠀⠀

What are Isotopes ?

⠀⠀⠀⠀

Atoms of same element having same atomic number but different atomic mass .

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀OR

Atom of same element having same number of protons and same number of electrons but having different number of neutrons

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What are Isobars ?

⠀⠀⠀⠀

Atoms of different elements having same mass number but different atomic number.

⠀⠀⠀⠀

What are Isotones ?

⠀⠀⠀⠀

Atoms of different elements having same number of neutrons are called isotones .

Answer 2

Answer:

Given that :-

⠀⠀⠀⠀

$\sf {\overline{\underline{ }}} X_1 \xrightarrow{\beta} {\overline{\underline{}}} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3 </p><p>$

X

1

β

X

2

α

75

238

X

3

⠀⠀⠀⠀

To Find :-

⠀⠀

Complete equation = ??

⠀⠀⠀⠀

Solution :-

⠀⠀⠀⠀

We know that atomic number is represented by Z and the mass no. is represented by A .

⠀⠀⠀⠀

Therefore ;

$\sf {}^A _ Z X_1 = {\overline{\underline{}}}X_1 </p><p>Z</p><p>A$

X

1

=

X

1

--- ( i )

⠀⠀⠀⠀

Due to the emission of a β - particle , the atomic number get increased by 1 , but the mass number remains unchanged

⠀⠀⠀⠀

$\sf {\overline{\underline{}}}X_2 = {} ^{A+1} _ {Z} X_2$

X

2

=

Z

A+1

X

2

--- ( ii )

⠀⠀⠀⠀

Due to the alpha decay the atomic number is decreased by 2 and the mass number is reduced by 4 .

⠀⠀⠀⠀

X

3

=

Z−2

A+1−4

X

3

--- ( iii )

⠀⠀⠀⠀

Now the equation becomes ;

⠀⠀⠀⠀

$\sf {} ^A _ Z X_1 \xrightarrow{\beta} {}^{A+1} _{Z} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3$

Z

A

X

1

β

Z

A+1

X

2

α

75

238

X

3

⠀⠀⠀⠀

⠀⠀⠀⠀

From eq ( iii )

⠀⠀⠀⠀

Firstly ; We observe that :-

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A + 1 - 4 = 238

⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A - 3 = 238

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A = 238 + 3

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀A = 241

⠀⠀⠀⠀

$\sf \: \: \: \: \: \: \:\: \: {\boxed{\large{A = 241}}} </p><p>A=241</p><p>$

⠀⠀⠀⠀

⠀⠀⠀⠀

Secondly ; we observe that :-

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀Z - 2 = 75

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀Z = 75 + 2

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀Z = 77

⠀⠀⠀⠀

⠀⠀⠀⠀

Now ; We can put the respective value in the final equation

⠀⠀⠀⠀

$\sf {} ^A _ Z X_1 \xrightarrow{\beta} {}^{A+1} _{Z} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3 </p><p>Z</p><p>A$

X

1

β

Z

A+1

X

2

α

75

238

X

3

⠀

$\sf {} ^{241} _ {77} X_1 \xrightarrow{\beta} {}^{241+1} _{77} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3 </p><p>77</p><p>241$

X

1

β

77

241+1

X

2

α

75

238

X

3

⠀⠀⠀⠀

$\sf {} ^{241} _{ 77 }X_1 \xrightarrow{\beta} {}^{242} _{77} X_2 \xrightarrow{\alpha} {}^{238}_{75} X_3 </p><p>77</p><p>241$

X

1

β

77

242

X

2

α

75

238

X

3

⠀⠀⠀⠀

⠀⠀⠀⠀

More information about the equation :-

⠀⠀⠀⠀

X₁ and X₂ are Isotopes

⠀⠀⠀⠀

Reason :- Becoz they have same atomic number but diff. mass number

⠀⠀⠀⠀

Extra Information

⠀⠀⠀⠀

What are Isotopes ?

⠀⠀⠀⠀

Atoms of same element having same atomic number but different atomic mass .

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀OR

Atom of same element having same number of protons and same number of electrons but having different number of neutrons

⠀⠀⠀⠀

What are Isobars ?

⠀⠀⠀⠀

Atoms of different elements having same mass number but different atomic number.

⠀⠀⠀⠀

What are Isotones ?

⠀⠀⠀⠀

Atoms of different elements having same number of neutrons are called isotones .