Home
Class 12
PHYSICS
Order of magnitude of density of uranium...

Order of magnitude of density of uranium nucleus is (`m_p=1.67xx10^-27kg`)

A

`10^(20) kg m^(-3)`

B

`10^(17) kg m^(-3)`

C

`10^(14) kg m^(-3)`

D

`10^(11) kg m^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
b
Nuclear density of an atom of mass numner A,
`D=(mass)/(volume)=(A(1.67 xx10^(-27)))/((4)/(3)pi[1.25 xx10^(15) A^(1//3)]^(3))`
`[:. V=4/3 pi R^(3), R=R_0A^(1//3), R_0=1.25 xx10^(-15)]`
`:. D=2xx10^(17) kg m^(-3)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

Order of magnitude of density of uranium nucleus is , [m = 1.67 xx 10^(-27 kg]

Order of magnitude of density of uranium nucleus is , [m = 1.67 x 10^(-27) kg]

The order of magnitude of the density of nuclear matter is=

The order of magnitude of the density of nuclear matter is 10^(4) kg m^(-3)

Assuming the radius of a nucleus to be equal to R=1.3 A^(1//3)xx10^(-15)m . Where A is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon =1.67xx10^(-27)kg

Assuming that protons and neutrons have equal masses, calculate how many times nuclear matter is denser than water. Take mass of a nucleon =1.67xx10^(-27)kg and R_(0)=1.2xx10^(-15)m .

The de Broglie wavelength of neutrons in thermal equilibrium is (given m_n=1.6xx10^(-27) kg)

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 xx10^(3) kg//m^(3) and its Young’s modulus is 9.27 xx10^(10) Pa. What will be the fundamental frequency of the longitudinal vibrations ?

A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm, then its kinetic energy (in MeV) is is (m_(p)=1.67xx10^(-27)kg, e=1.6xx10^(-19)C)

Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in .^(1)H and .^(2)H . This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass mu , revolving around the nucleus at a distance equal to the electron -nucleus separation. Here mu = m_(e) M//(m_(e)+M) , where M is the nuclear mass and m_(e) is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in .^(1)H and .^(2)H . (mass of .^(1)H nucleus is 1.6725 xx 10^(-27) kg, mass of .^(2)H nucleus is 3.3374 xx 10^(-27) kg, Mass of electron = 9.109 xx 10^(-31) kg .)