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shusha [124]
4 years ago
15

Use the helium-4 isotope to define atomic number and mass number. Why does knowledge of the atomic number enable us to deduce th

e number of electrons present in an atom?
Chemistry
1 answer:
Zanzabum4 years ago
3 0

<u>Answer:</u> The atomic number of helium atom is 2 and mass number of helium atom is 4.

<u>Explanation:</u>

Atomic number is defined as the number of protons or electrons that are present in a neutral atom.  It is represented by the symbol 'Z'

Atomic number = number of protons = number of electrons

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.  It is represented by the symbol 'A'

Mass number = Number of protons + Number of neutrons

The general representation of an isotope is given as _Z^A\textrm{X}

where,

A = mass number of element

Z = atomic number of element

X = symbol of the element

We are given:

An isotope of helium having formula of _2^4\textrm{He}

where,

Atomic number of helium = 2 = Number of protons = Number of electrons

Mass number of helium = Number of protons + Number of neutrons

4 = 2 + Number of neutrons

Number of neutrons = 4 - 2 = 2

Hence, the atomic number of helium atom is 2 and mass number of helium atom is 4.

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Answer: (a) BE = 1.112 MeV

               (b) BE = 7.074 MeV

               (c) BE = 7.767 MeV

               (d) BE = 8.112 MeV

Explanation: Binding energy per nucleon is the average energy necessary to remove a proton or a neutron from the nucleus of an atom. It is mathematically defined as:

BE = \frac{\Delta m.c^{2}}{A}

Where

Δm is a difference in mass known as <u><em>mass</em></u><em> </em><u><em>defect</em></u>

A is atomic mass of an atom.

Mass Defect is determined by:

\Delta m =Zm_{p}+(A-Z)m_{n} - m_{nuc}

where:

Z is atomic number

m_{p} is mass of proton

m_{n} is mass of neutron

m_{nuc} is mass of the nucleus

Mass of proton is 1.007825u.

Mass of neutron is 1.008665u.

The unit u is equal to 931.5MeV/c².

(a) 2H(deuterion): Given: Z = 1; A = 2; m_{nuc} = 2.014102u

\Delta m =1(1.007825)+1(1.008665) -2.014102

\Delta m =0.002388u

BE = \frac{0.002388.c^{2}}{2}.931.5\frac{MeV}{c^{2}}

BE = 1.112MeV

(b) 4He (Helium): Given: Z = 2; A = 4; m_{nuc} = 4.002603

\Delta m =2(1.007825)+2(1.008665) -4.002603

\Delta m =0.030377u

BE = \frac{0.030377.c^{2}}{4}.931.5\frac{MeV}{c^{2}}

BE = 7.074MeV

(c) 18O (Oxygen): Given: Z = 8; A = 18; m_{nuc} = 17.999160

\Delta m =8(1.007825)+10(1.008665) -17.999160

\Delta m =0.15009u

BE = \frac{0.15009.c^{2}}{18}.931.5\frac{MeV}{c^{2}}

BE = 7.767MeV

(d) 23Na (Sodium): Given: Z = 11; A = 23; m_{nuc} = 22.989767

\Delta m =11(1.007825)+12(1.008665) -22.989767

\Delta m =0.200288u

BE = \frac{0.200288.c^{2}}{23}.931.5\frac{MeV}{c^{2}}

BE = 8.112MeV

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