Q.44. The binding energy per nucleon of is __________ .
[Take , , , ]

Q: What is the correct answer for this exam Physics Q44?

The correct answer is B. For ${}_{83}^{209}\mathrm{Bi}$, number of protons $Z = 83$, number of neutrons $N = A - Z = 209 - 83 = 126$. Mass defect $\Delta m = (Z m_p + N m_n) - m_{\text{nucleus}}$. $\Delta m = (83 \times 1.007825 \mathrm{~u} + 126 \times 1.008665 \mathrm{~u}) - 208.980388 \mathrm{~u}$. $\Delta m = (83.649475 \mathrm{~u} + 127.09179 \mathrm{~u}) - 208.980388 \mathrm{~u}$. $\Delta m = 210.741265 \mathrm{~u}

Q: Q.44. The binding energy per nucleon of is __________ . [Take , , , ]

The correct answer is B. For ${}_{83}^{209}\mathrm{Bi}$, number of protons $Z = 83$, number of neutrons $N = A - Z = 209 - 83 = 126$. Mass defect $\Delta m = (Z m_p + N m_n) - m_{\text{nucleus}}$. $\Delta m = (83 \times 1.007825 \mathrm{~u} + 126 \times 1.008665 \mathrm{~u}) - 208.980388 \mathrm{~u}$. $\Delta m = (83.649475 \mathrm{~u} + 127.09179 \mathrm{~u}) - 208.980388 \mathrm{~u}$. $\Delta m = 210.741265 \mathrm{~u}

Q: What are the options for this this exam Physics question?

A. 48 · B. 84 · C. 79 · D. 94

Question

Q.44. The binding energy per nucleon of  is __________ .
[Take , , , ]

FirstInTest · Accepted Answer

For ${}_{83}^{209}\mathrm{Bi}$, number of protons $Z = 83$, number of neutrons $N = A - Z = 209 - 83 = 126$.
Mass defect $\Delta m = (Z m_p + N m_n) - m_{	ext{nucleus}}$.
$\Delta m = (83 	imes 1.007825 \mathrm{~u} + 126 	imes 1.008665 \mathrm{~u}) - 208.980388 \mathrm{~u}$.
$\Delta m = (83.649475 \mathrm{~u} + 127.09179 \mathrm{~u}) - 208.980388 \mathrm{~u}$.
$\Delta m = 210.741265 \mathrm{~u} - 208.980388 \mathrm{~u} = 1.760877 \mathrm{~u}$.
Binding Energy (BE) $= \Delta m 	imes 931 \mathrm{~MeV/u}$.
BE $= 1.760877 	imes 931 \mathrm{~MeV} = 1639.3764 \mathrm{~MeV}$.
Binding Energy per nucleon (BE/A) $= \frac{1639.3764 \mathrm{~MeV}}{209} \approx 7.8439 \mathrm{~MeV}$.

Answer

48

Answer

84

Answer

79

Answer

94

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