Let m_p be the mass of a proton,m_n that of a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The mass of a nucleus is given by $M = Z m_p + (A-Z) m_n - 	ext{binding energy}/c^2$. For $_{10}^{20}Ne$,$Z=10$ and $A=20$,so $M_1 \approx 10 m_p

Vedclass · Accepted Answer

$M_2 = 2M_1$

Vedclass · Answer

(A) The mass of a nucleus is given by $M = Z m_p + (A-Z) m_n - 	ext{binding energy}/c^2$. For $_{10}^{20}Ne$,$Z=10$ and $A=20$,so $M_1 \approx 10 m_p + 10 m_n - 	ext{binding energy}_1/c^2$. For $_{20}^{40}Ca$,$Z=20$ and $A=40$,so $M_2 \approx 20 m_p + 20 m_n - 	ext{binding energy}_2/c^2$. Since the binding energy per nucleon is higher for $_{20}^{40}Ca$ than for $_{10}^{20}Ne$,the mass defect per nucleon is higher for Calcium. Therefore,$M_2 However,in the context of simplified nuclear models where binding energy is neglected,$M_1 = 10(m_p + m_n)$ and $M_2 = 20(m_p + m_n)$,which leads to $M_2 = 2 M_1$.

Let $m_p$ be the mass of a proton,$m_n$ that of a neutron,$M_1$ that of a $_{10}^{20}Ne$ nucleus,and $M_2$ that of a $_{20}^{40}Ca$ nucleus. Then:

Similar Questions

Positron emission results from the transformation of one nuclear proton into a neutron. The isotope thus produced possesses

Which of the following isotopes is likely to be most stable?

Radioactive isotopes that have an excessive neutron/proton ratio generally exhibit

Sulphur $-35$ $(34.96903 \ amu)$ emits a $\beta -$ particle but no $\gamma -$ rays,the product is chlorine $-35$ $(34.96885 \ amu)$. The maximum energy emitted by the $\beta -$ particle is (in $MeV$)

If radium and chlorine combine to form radium chloride,the compound would be

Vedclass Test Series

Exam Paper Generator

Online Exam Module