Question

An ion has one neutron more than the number of protons but has one electron less than the number of protons. Which two ions are possible which satisfy this condition? Explain.

Answer 1

Answer:

Correct Question -

The circumference of two circle are in the ratio 2 : 3. Find the ratio of their areas.

Given -

Ratio of their circumference = 2:3

To find -

Ratio of their areas.

Formula used -

Circumference of circle

Area of circle.

Solution -

In the question, we are provided, with the ratios of the circumference of 2 circles, and we need to find the ratio of area of those circle, for that first we will use the formula of circumference of a circle, then we will use the formula of area of circles. We will be writing 1 equation in it too.

So -

Let the circumference of 2 circles be c1 and c2

According to question -

c1 : c2

Circumference of circle = 2πr

where -

π = $\tt\dfrac{22}{7}$

r = radius

On substituting the values -

c1 : c2 = 2 : 3

2πr1 : 2πr2 = 2 : 3

$\tt\dfrac{2\pi\:r\:1}{2\pi\:r\:2}$ = $\tt\dfrac{2}{3}$

$\tt\dfrac{r1}{r2}$ = $\tt\dfrac{2}{3}$$\longrightarrow$ [Equation 1]

Now -

Let the areas of both the circles be A1 and A2

Area of circle = πr²

So -

Area of both circles = πr1² : πr2²

On substituting the values -

A1 : A2 = πr1² : πr2²

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(\pi\:r1)}{(\pi\:r2)}^{2}$

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(r1)}{(r2)}^{2}$

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(2)}{(3)}^{2}$ [From equation 1]

So -

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{4}{9}$

$\therefore$ The ratio of their areas is 4 : 9

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Answer 2

Explanation:

The most abundant isotope, hydrogen-1, protium, or light hydrogen, contains no neutron. Thus, the answer would be false. But, isotope of hydrogen namely, Deuterium, has 1 proton, 1 electron, and 1 neutron. Atoms of the same element with different numbers of neutrons are called isotopes.

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