Home
Class 12
PHYSICS
The energy required to separate the typi...

The energy required to separate the typical middle mass nucleus `._(50)^(120)Sn` into its constituent nucleons ( Mass of `._(50)^(120)sn=119.902199u`, mass of proton `=1.007825 u` and mass of neutron `=1.008665 u`)

A

951 MeV

B

805 MeV

C

1021 MeV

D

1212 MeV

Text Solution

Verified by Experts

The correct Answer is:
C

`Delta E = Delta m 931.5 MeV`
Promotional Banner

Topper's Solved these Questions

  • NUCLEI

    NARAYNA|Exercise EXERCISE-2 (C.W)|14 Videos
  • NUCLEI

    NARAYNA|Exercise EXERCISE-2 (H.W)|9 Videos
  • NUCLEI

    NARAYNA|Exercise EXERCISE-1 (C.W)|21 Videos
  • NUCLEAR PHYSICS

    NARAYNA|Exercise LEVEL-II-(H.W)|9 Videos
  • RAY OPTICS AND OPTICAL INSTRAUMENTS

    NARAYNA|Exercise EXERCISE- 4 One or more than one correct answer type|13 Videos

Similar Questions

Explore conceptually related problems

The binding energy per nucleon of ""_(7)N^(14) nucleus is: (Mass of ""_(7)N^(14) = 14.00307 u ) mass of proton = 1.007825 u mass of neutron = 1.008665 u

The binding energy of ._(10)Ne^(20) is 160.6 MeV. Find its atomic mass. Take mass of proton =1.007825 u and mass of neutron =1.008665 u.

What is the binding energy per nucleon in ._2He^4 Given , Mass of ._2He^4 =4.002604 amu Mass of proton =1.007825 amu Mass of neutron =1.008665 amu

Calculate the binding energy per nucleon of ._26Fe^(56) nucleus. Given that mass of ._26Fe^(56)=55.934939u , mass of proton =1.007825u and mass of neutron =1.008665 u and 1u=931MeV .

Calculate the binding energy per nucleon of ._(20)^(40)Ca . Given that mass of ._(20)^(40)Ca nucleus = 39.962589 u , mass of proton = 1.007825 u . Mass of Neutron = 1.008665 u and 1 u is equivalent to 931 MeV .

Find energy required to break an aluminium nucleus into its constituent nucleons ( m_n = 1.00867 u , m_p = 1.00783 u , m_Al = 26.98154 u )

Find the binding energy of ._(20)^(56)Fe . Atomic mass of .^(56)Fe is 55.934939 amu. Mass of a proton is 1.007825 amu and that of neutron=1.008665 amu.

What energy is needed to split an alpha particle to piece? Given mass of alpha particle is 4.0039a.m.u, mass of proton=1.007825u and mass of neutron =1.008665u use 1a.m.u.=931MeV