Table gives the masses (in u) of several particles. Avogadro constant N A = 6 . 02 × 10 23 mol - 1 . Use the table to determine to three significant figures: (a) the mass in $\mathrm{kg}$ of a helium- 4 nucleus.

Given, m a s s o f H 2 4 e = 4 . 002602 u Now, ∵ 1 u = 1 . 6605 × 10 - 27 Kg ∴ Mass of H 2 4 e = 4 . 002602 × 1 . 6605 × 10 - 27 Kg ⇒ Mass of H 2 4 e ≈ 6 . 65 × 10 - 27 Kg

Table gives the masses (in u) of several particles. Avogadro constant N A = 6 . 02 × 10 23 mol - 1 . Use the table to determine to three significant figures: (a) the mass in $\mathrm{kg}$ of a helium- 4 nucleus.

Given, m a s s o f H 2 4 e = 4 . 002602 u Now, ∵ 1 u = 1 . 6605 × 10 - 27 Kg ∴ Mass of H 2 4 e = 4 . 002602 × 1 . 6605 × 10 - 27 Kg ⇒ Mass of H 2 4 e ≈ 6 . 65 × 10 - 27 Kg

Table gives the masses (in u) of several particles. Avogadro constant N A = 6 . 02 × 10 23 mol - 1 . Use the table to determine to three significant figures: (b) the mass in gram (g) of $1.0$ mole of uranium-235 nuclei.

One mole of any substance has same mass in grams as it has in amu ∴ m = 235 . 043930 g

(a) Explain why hydrogen H 1 1 (proton) cannot appear on the graph shown in Figure.

Nucleus of hydrogen has only one proton. Hence its binding energy will be zero . The nuclear binding energy is the energy it takes to separate all the nucleons in a nucleus from each other. Since there is only the one nucleon, it's already separated from any other nucleons. Therefore, hydrogen is not shown in the given graph.

The mass of a 4 8 Be nucleus is 1 . 33 × 10 - 26 kg . For the nucleus of Be 4 8 , determine: (a) the mass defect in $\mathrm{kg}$

Using the given values in book we have, M a s s d e f e c t Δ m = 4 m p + 4 m n – m B e ⇒ ∆ m ≈ ( 8 × 1 . 67 – 13 . 3 ) × 10 - 27 ⇒ ∆ m = 6 . 00 × 10 - 29 kg

The mass of a 4 8 Be nucleus is 1 . 33 × 10 - 26 kg . For the nucleus of Be 4 8 , determine: (b) the binding energy of the nucleus in $\mathrm{MeV}$

Using the given values in book we have, M a s s d e f e c t Δ m = 4 m p + 4 m n – m B e ⇒ ∆ m ≈ ( 8 × 1 . 67 – 13 . 3 ) × 10 - 27 ⇒ ∆ m = 6 . 00 × 10 - 29 kg Now, B i n d i n g e n e r g y E B = Δ m c 2 ⇒ E B = 6 . 00 × 10 - 29 × ( 3 . 00 × 10 8 ) 2 ⇒ E B = 5 . 4 × 10 - 12 J ⇒ E B = 5 . 4 × 10 - 12 1 . 6 × 10 - 19 e V = 3 . 4 × 10 7 e V ⇒ E B = 34 Mev

The mass of a 4 8 Be nucleus is 1 . 33 × 10 - 26 kg . For the nucleus of Be 4 8 , determine: (c) the binding energy (in $\mathrm{MeV}$ ) per nucleon for the nucleus.

Using the given values in book we have, M a s s d e f e c t Δ m = 4 m p + 4 m n – m B e ⇒ ∆ m ≈ ( 8 × 1 . 67 – 13 . 3 ) × 10 - 27 ⇒ ∆ m = 6 . 00 × 10 - 29 kg Now, B i n d i n g e n e r g y E B = Δ m c 2 ⇒ E B = 6 . 00 × 10 - 29 × ( 3 . 00 × 10 8 ) 2 ⇒ E B = 5 . 4 × 10 - 12 J ⇒ E B = 5 . 4 × 10 - 12 1 . 6 × 10 - 19 e V = 3 . 4 × 10 7 e V ⇒ E B = 34 Mev Since there are total 8 nucleons therefore, B i n d i n g e n e r g y p e r n u c l e o n = E B n ⇒ E B n = 34 Mev 8 = 4 . 25 Mev

EASY

AS and A Level

IMPORTANT

Earn 100

Table gives the masses (in u) of several particles. $(Avogadro constant N_{A} = 6.02 \times 10^{23} {mol}^{- 1})$ . Use the table to determine to three significant figures:

(a) the mass in $\mathrm{kg}$ of a helium- 4 nucleus.

Important Questions on Nuclear Physics

EASY

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>Questions>Q 7

Table gives the masses (in u) of several particles. $(Avogadro constant N_{A} = 6.02 \times 10^{23} {mol}^{- 1})$ . Use the table to determine to three significant figures: