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Assign groups...

Question

Assign groups to the elements A, B, C, D and E starting from element X(group II A)

$X (I I A) - α ⟶ A - β ⟶ B - β ⟶ C - α ⟶ D$

The relation between A and C is:

A

A and C are isotopes.

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B

A and C are isobars.

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C

A and C are both isobars as well as isotopes.

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D

None of these.

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Solution

The correct option is C A and C are isobars.
$α$ contains 2 protons and 2 neutrons. Mass no. 4
$β$ contains 1 electron. No mass no.

Let X have $x$ and $y$ as atomic no. and mass no. respectively.

$X_{x}^{y} (I I A) - α ⟶ A_{x}^{y - 4} - β ⟶ B_{x - 1}^{y - 4} - β ⟶ C_{x - 2}^{y - 4} - α ⟶ D_{x - 2}^{y - 8}$

Both A and C have the same mass no. so they are isobars.

Hence, the correct option is B

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Similar questions

A, B

C

are isodiaphers while

C, D

E

are isobars. Calculate the difference of protons between

A

E

_{82}^{206} A ⟶ B ⟶ C ⟶ D ⟶ E .

Given: Isodiaphers and isobars are formed in successive

α

β -

emission respectively.

Q. Which of the given relations about the below elements are correct.

a)^{12} C b)^{14} C c)^{14} N d)^{16} O

\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}

\frac{x - b + c}{β - r} = \frac{y - b}{β} = \frac{z - b - c}{β + r}

are coplanar and then equation to the plane in which they lie, is?

α, β

are the roots of

a x^{2} + b x + c = 0

then match the elements of list I with elements of list II:

List I	List II
A) $\frac{α}{β} + \frac{β}{α} =$	1) $\frac{c^{2}}{a^{2}}$
B) $\frac{α^{2} + β^{2}}{α^{- 2} + β^{- 2}} =$	2) $\frac{c^{5} [3 a b c - b^{3}]}{a^{8}}$
C) $α^{3} + β^{3}$	3) $\frac{b^{2} - 2 a c}{a c}$
D) $α^{5} β^{8} + α^{8} β^{5} =$	4) $\frac{3 a b c - b^{3}}{a^{3}}$

The correct match from List-I to List-II:

A, B, C, D

Q. Show that the lines

\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}

\frac{x - b + c}{β - γ} = \frac{y - b}{β} = \frac{z - b - c}{β + γ}

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