KVPY SX SB Exam 20141 Question Paper

Question : 40 of 120

Marks: +1, -0

The binding energy per nucleon of

_5 B ^10

is 8.0 MeV and that of

_5B ^11

is 7.5 MeV. The energy required to remove a neutron from

_5B ^11

is (mass of electron and proton are

9.11 ×10 ^{-31}

kg and

1.67 ×10^{-27 }

kg, respectively)

Solution:

formation of

_5B^10

will release energy ⇒

E_2

E_1

= Binding energy of

_5B^11 ⇒ 7.5 × 11

MeV
= 82.5 MeV

E_2 =

Binding energy of

_5B^10 = 8.0 × 10

= 80 MeV
Energy given

= E_1 -E_2

= 82.5 – 80
= 2.5 MeV

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