The masses of neutron,proton and deuteron in | vedclass

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

Question

(A) Given: Mass of neutron $m_n = 1.00893 \ amu$,mass of proton $m_p = 1.00813 \ amu$,and mass of deuteron $m_d = 2.01473 \ amu$. $A$ deuteron nucleus

Vedclass · Accepted Answer

$11.65 	imes 10^{-4}$

Vedclass · Answer

(A) Given: Mass of neutron $m_n = 1.00893 \ amu$,mass of proton $m_p = 1.00813 \ amu$,and mass of deuteron $m_d = 2.01473 \ amu$. $A$ deuteron nucleus $(^2_1H)$ consists of one proton and one neutron. The mass defect $\Delta m$ is calculated as: $\Delta m = (m_n + m_p) - m_d$ $\Delta m = (1.00893 + 1.00813) - 2.01473$ $\Delta m = 2.01706 - 2.01473 = 0.00233 \ amu$. The packing fraction is defined as the ratio of mass defect to the mass number $(A)$. For deuteron,$A = 2$. $	ext{Packing fraction} = \frac{\Delta m}{A} = \frac{0.00233}{2} = 0.001165 = 11.65 	imes 10^{-4}$.

The masses of neutron,proton and deuteron in amu are $1.00893$,$1.00813$ and $2.01473$,respectively. The packing fraction of the deuteron in amu is

Similar Questions

In the $C.G.S.$ system,the magnitude of the force is $100 \ dynes$. In another system where the fundamental physical quantities are $kilogram$,$metre$,and $minute$,the magnitude of the force is:

The locus of the centroid of the triangle with vertices at $(a \cos \theta, a \sin \theta)$,$(b \sin \theta, -b \cos \theta)$ and $(1, 0)$ is (where $\theta$ is a parameter).

Which is the best method for the conservation of wild relatives of plants?

If a curve $y = a\sqrt{x} + bx$ passes through the point $(1, 2)$ and the area bounded by the curve,the line $x = 4$,and the $x$-axis is $8$ sq. units,then:

For intensity $I$ of a light of wavelength $5000 \mathring{A}$,the photoelectron saturation current is $0.40 \mu A$ and the stopping potential is $1.36 V$. The work function of the metal is ............. $eV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module