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Standard XII

Physics

Mass Energy Equivalence

The atomic ma...

Question

The atomic mass of $7^{N^{15}}$ is 15.000108 a.m.u. and that is of $8^{◯^{16}}$ 15.994915 a.m.u. If the mass of a proton is 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is

0.0130181 MeV

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12.13 MeV

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13.018 MeV

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12.13 eV

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Solution

The correct option is B 12.13 MeV
$\begin{matrix} (M_{n} + M_{H} - M_{0}) \times 431.5 = [15.000108 + 1.007827 - 15.994915] \times 431 = 12.13 M e v \end{matrix}$

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Similar questions

Q. The atomic mass of

_{7} N^{15}

15.000108 a m u

and that of

_{8} O^{16}

15.994915 a m u

. The minimum energy required to remove the least tightly bound proton is (mass of proton is

1.007825 a m u)

Q. The masses of neutron and proton are

1.0087 a . m . u .

and

1.0073 a . m . u .

respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass

4.0015 a . m . u

, then the binding energy fo the helium nucleus will be: (

1 a . m . u . = 931 M e V

)

Q. Calculate the binding energy of an alpha-particle. Given mass of proton

= 1.0073

a.m.u., mass of neutron

= 1.0087

a.m.u., mass of

α

-particle

= 4.0015

a.m.u.

Q. The mass of deutron

(_{1} H^{2})

nucleus is

2.013553

a.m.u. If the masses of proton and neutron are

1.007275

a.m.u. and

1.008665

a.m.u. respectively. Calculate the mass defect, the packing fraction, binding energy and binding energy per nucleon.

Q. Obtain the binding energy of the nuclei

_{26} F e^{56}

and

_{83} B i^{209}

in units of MeV from the following data: mass of hydrogen atom

= 1.007825

a.m.u., mass of neutron

= 1.08665

a.m.u.; mass of

_{26} F e^{56}

atom

= 55.934939

a.m.u. and mass of

_{83} B i^{209}

atom

= 208.980388

a.m.u. Also calculate binding energy per nucleon in the two cases.

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The atomic mass of 7N15 is 15.000108 a.m.u. and that is of 8◯16 15.994915 a.m.u. If the mass of a proton is 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is

The correct option is B 12.13 MeV(Mn+MH−M0)×431.5=[15.000108+1.007827−15.994915]×431=12.13Mev

The atomic mass of $7^{N^{15}}$ is 15.000108 a.m.u. and that is of $8^{◯^{16}}$ 15.994915 a.m.u. If the mass of a proton is 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is

The correct option is B 12.13 MeV
$\begin{matrix} (M_{n} + M_{H} - M_{0}) \times 431.5 = [15.000108 + 1.007827 - 15.994915] \times 431 = 12.13 M e v \end{matrix}$