Ncert Exercises & Solutions

A nuclide 1 is said to be the mirror isobar of nuclide 2 if Z₁ = N₂ and Z₂ = N₁. (a) What nuclide is a mirror isobar of $_{11}^{23} Na$ $_{11}^{23} Na$ ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?

:NEETprep Answer:

Hint: The binding energy of a nucleus depends on the no. of nucleons.

Step 1: Find the mirror isobars.

(a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2, if Z₁ = N₂ and Z₂ = N₁

Now in_{11} {Na}^{23}, Z_{1} = 11, N_{1} = 23 - 11 = 12

$Now in_{11} {Na}^{23}, Z_{1} = 11, N_{1} = 23 - 11 = 12$

∴ Mirror isobar of_{11} {Na}^{23} is_{12} {Mg}^{23}, for which Z_{2} = 12 = N_{1} and N_{2} = 23 - 12 = 11 = Z_{1}

$∴ Mirror isobar of_{11} {Na}^{23} is_{12} {Mg}^{23}, for which Z_{2} = 12 = N_{1} and N_{2} = 23 - 12 = 11 = Z_{1}$

Step 2: Find the no. of nucleons in each isobar.

(b) As

_{12}^{23} Mg

$_{12}^{23} Mg$ contains an even number of protons (12) against

_{11}^{23} Na

$_{11}^{23} Na$ which has an odd number of protons (11), therefore

_{12}^{23} Mg

$_{12}^{23} Mg$ has greater binding energy than

_{11} {Na}^{23}

$_{11} {Na}^{23}$ .

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