Home
Class 12
PHYSICS
Let m(p) be the mass of a poton , M(1) t...

Let `m_(p)` be the mass of a poton , `M_(1)` the mass of a `_(10)^(20) Ne` nucleus and `M_(2)` the mass of a `_(20)^(40) Ca` nucleus . Then

A

`M_(2) = 2 M_(1)`

B

`M_(2) gt 2M_(1)`

C

`M_(2) lt 2M_(1)`

D

`M_(1) lt 10(m_(n) + m_(p))`

Text Solution

Verified by Experts

The correct Answer is:
C, D
Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles. `._(1.0)^(20)Ne` is made up of `10` protons plus `10` neutrons.
Therefore, mass of `._(10)^(20)Ne` nucleus,
`M_(1) lt 10 (m_(p)+m_(n))`
Also, heavier the nucleus, more is the mass defect. Thus, `20 (m_(n)+m_(p))-M_(2) gt 10 (m_(p)+m_(n))-M_(1)`
`implies 10 (m_(p)+m_(n)) gt M_(2)-M_(1) implies M_(2) lt M_(1)+10(m_(p)+m_(n))`
Now since `M_(1) lt 10 (m_(p)+m_(n)) :. M_(2) lt 2M_(1)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Match The Column|9 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Comprehension #1|3 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Exercise-05(B)|46 Videos

  • RACE

    ALLEN |Exercise Basic Maths (Wave Motion & Dopplers Effect) (Stationary waves & doppler effect, beats)|25 Videos

  • TEST PAPER

    ALLEN |Exercise PHYSICS|4 Videos

Similar Questions

Explore conceptually related problems

How mass of atom or nucleus is determined ?

If M_(e) is the mass of earth and M_(m) is the mass of moon (M_(e)=81 M_(m)) . The potential energy of an object of mass m situated at a distance R from the centre of earth and r from the centre of moon, will be :-

Knowledge Check

  • In a Rutherford's scattering experiment when a projectile of chargez Z_(1) and mass M_(1) approaches a target nucleus of Z_(2) and mass M_(2) the distance of closest approach is r_(0) . The energy of the projectile is........

    A
    directly proportional to `Z_(1), Z_(2)`
    B
    inversely proportional to `Z_(1)`
    C
    directly proportional to `M_(1)`
    D
    directly proportional to `M_(1),M_(2)`

  • A nucleus of mass M is initially at rest.It absorb neutron having m_(N) rest mass compound nucleus divided in two nucleus of m_(1) and 5 m_(1) .If wavelength of nucleus with mass m_(1) is lambda ,then wavelength of other nucleus will be…..

    A
    `25 lambda`
    B
    `5 lambda`
    C
    `(lambda)/(5)`
    D
    `lambda`

  • Mass of photon moving with speed v is ……, m_(0) is rest mass of photon.

    A
    zero
    B
    `(m_(0 ))/(sqrt(1-(c^(2))/(v^(2))))`
    C
    `(m_(0))/(sqrt(1-(v^(2))/c^(2)))`
    D
    infinite

    • Similar Questions

    Explore conceptually related problems

    If M_(o) is the mass of an oxygen isotope ._(8)O^(17), M_(p) and M_(N) are the masses of a proton and neutron respectively, the nuclear binding energy of the isotope is:

    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 u, 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 =9.709xx10^(-31)kg)

    The mass of nucleus ._(z)X^(A) is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of mass m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of massws m_(3) and m_(4) can undergo complete fusion and form a heavy nucleus of mass M ''only if (m_(3)+m_(4)) gt M ''. The masses of some neutral atoms are given in the table below. |{:(._(1)^(1)H,1.007825u,._(1)^(2)H,2.014102u,),(._(1)^(3)H,3.016050u,._(2)^(4)H,4.002603u,),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,),(._(30)^(70)Zn,69.925325u,._(34)^(82)Se,81.916709u,),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.97445u,),(._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u,):}| The correct statement is

    The mass of a nucleus ._(Z)^(A)X is less that the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus.The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of masses m_(3) and m_(4) can undergo complete fusion and form a heavy nucleus of mass M'. only if (m_(3)+m_(4)) gt M' . The masses of some neutral atoms are given in the table below: |{:(._(1)^(1)H ,1.007825u , ._(1)^(2)H,2.014102u,._(1)^(3)H,3.016050u,._(2)^(4)He,4.002603u),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,._(30)^(70)Zn,69.925325u, ._(34)^(82)Se,81.916709u),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.974455u,._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u):}| Taking kinetic energy ( in KeV ) of the alpha particle, when the nucleus ._(84)^(210)P_(0) at rest undergoes alpha decay, is:

    Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take molar mass of helium M_(1) = 4g and that of hydrogen M_(2) = 2g The equivalent molar mass of the mixture is

    Home
    Profile