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Standard IX

Chemistry

Atomic Number

M, M n and ...

Question

$M, M$ $_{n}$ and $M$ $_{p}$ denotes the masses of a nucleus of $_{Z} X^{A}$ , a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

M = (A - Z) M_{n} + Z M_{p}

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M = Z M_{n} + (A - Z) M_{p}

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M > (A - Z) M_{n} + Z M_{p}

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M < (A - Z) M_{n} + Z M_{p}

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Solution

The correct option is D $M < (A - Z) M_{n} + Z M_{p}$
$M c^{2} + B i n d i n g e n e r g y = [(A - Z) M n + Z M p] c^{2}$
Therefore, mass of nucleus is less than total mass of its free nucleons.

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Similar questions

Q. If m,

m_{n}

and

m_{p}

are the masses of

_{Z} X^{A}

nucleus, neutron and proton respectively, then which of the following is correct?

If m, $m_{n}$ and $m_{p}$ are the masses of $_{Z} X^{A}$ nucleus, neutron and proton respectively, then which of the following is correct?

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represent mass of neutron and proton respectively. If an element having nuclear mass M has N-neutron and Z-proton, then the correct relation will be

$M_{n}$ and $M_{p}$ represent mass of neutron and proton respectively. If an element having nuclear mass M has N-neutron and Z-proton, then the correct relation will be

Q. Nuclear binding energy is the energy released during the hypothetical formation of the nucleus by the condensation of individual nucleons. Thus, binding energy per nucleon =

\frac{Total binding energy}{Number of nucleons}

For example, the mass of hydrogen atom is equal to the sum of the masses of a proton and an electron. For other atoms, the atomic mass is less than the sum of the masses of protons, neutrons and electrons present. This difference in mass, termed as mass defect, is a measure of the binding energy of protons and neutrons in the nucleus. The mass-energy relationship postulated by Einstein is expressed as:

Δ E = Δ m c^{2}

Where

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M,Mn and Mp denotes the masses of a nucleus of ZXA, a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

The correct option is D M<(A−Z)Mn+ZMpMc2+Bindingenergy=[(A−Z)Mn+ZMp]c2Therefore, mass of nucleus is less than total mass of its free nucleons.

$M, M$ $_{n}$ and $M$ $_{p}$ denotes the masses of a nucleus of $_{Z} X^{A}$ , a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

The correct option is D $M < (A - Z) M_{n} + Z M_{p}$
$M c^{2} + B i n d i n g e n e r g y = [(A - Z) M n + Z M p] c^{2}$
Therefore, mass of nucleus is less than total mass of its free nucleons.