Advertisements
Advertisements
प्रश्न
Answer the following question.
Show that the density of the nucleus is independent of its mass number A.
Show that nuclear density in a given nucleus is independent of mass number A.
Advertisements
उत्तर १
Density of the nucleus = `(\text { mass of nucleus})/(\text { volume of nucleus})`
Mass of the nucleus = A amu = A × 1.66 × 10-27 kg
Volume of the nucleus = `4/3πR^3 = 4/3π(R_0A^(1/3))^3 A =4/3πR_0^3 A`
Thus, density = `(Axx1.66 xx10^(-27))/((4/3πR_0^3)A)` = `(1.66 xx 10^-27)/((4/3πR_0^3)`
which shows that the density is independent of mass number A.
Using R0 = 1.1 × 10-15 m and density = 2.97 × 1017 kg m-3
उत्तर २
Density = `("p") = "m"/"v"`
R = `"R"_0("A")^(1/3)`
Volume = `(4)/(3)"πR"^3`
Volume of atom = `(4)/(3)π ("R"_0 "A"^(1/3))^3`
= `(4)/(3)π"R"_0^3"A"`
Density = `"mass"/"volume"`
`"p" = ("m"."A")/((4)/(3) "πR"_0^3 "A")`
`"p" = ("m")/((4)/(3) "πR"_0^3)`
m = mass of proton
It shows independence in mass.
APPEARS IN
संबंधित प्रश्न
A molecule of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behaviour of a hydrogen molecule. Why?
The radius of \[\ce{^27_13X}\] nucleus is R. The radius of \[\ce{^125_53Y}\] nucleus will be ______.
The radius of inner most orbit of hydrogen atom is 5.3 × 10-11 m. What is the radius of third allowed orbit of hydrogen atom?
Assertion (A): All nuclei are not of the same size.
Reason (R): The size of the nucleus depends on atomic mass.
How is the nuclear radius defined?
Which experiments provide more accurate measurements of nuclear size and help determine the distribution of charge inside the nucleus?
If the mass number \[A\] of a nucleus increases, its radius increases as \[A^{1/3}\]. This implies that:
Since the nucleus is approximately spherical with \[R = R_0 A^{1/3}\], its volume \[V = \frac{4}{3}\pi R^3\] is proportional to:
The approximate density of nuclear matter is:
For an iron nucleus with mass number \[A = 56\] and mass \[55.85\,\text{u}\], taking \[R_0 = 1.2 \times 10^{-15}\,\text{m}\] and \[R = R_0 A^{1/3}\], the calculated nuclear density using \[\rho = \frac{m}{V}\] is about:
