Draw atomic depictions similar to those in Problem 2.43 for $\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}{}_{\mathbf{82}}{}^{\mathbf{207}}\mathit{P}\mathit{b}\mathbf{}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}{}_{\mathbf{5}}{}^{\mathbf{9}}\mathit{B}\mathit{e}\mathbf{}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}{}_{\mathbf{33}}{}^{\mathbf{75}}\mathit{A}\mathit{s}$

${}_{82}{}^{207}Pb$

Given,

Atomic number = 82

Atomic mass = 207

Number of protons = Atomic number = 82

Number of electrons = 82

We know,

Atomic mass = number of protons + number of neutrons

So,

Number of neutrons = atomic mass – number of protons

Number of neutrons = 207 – 82 = 125

${}_5{}^{9}Be$

Given,

Atomic number = 5

Atomic mass = 9

Number of protons = Atomic number = 5

Number of electrons = 5

We know,

Atomic mass = number of protons + number of neutrons

So,

Number of neutrons = atomic mass – number of protons

Number of neutrons = 9 – 5 = 4

${}_{33}{}^{75}As$

Given,

Atomic number = 33

Atomic mass = 75

Number of protons = Atomic number = 33

Number of electrons = 33

We know,

Atomic mass = number of protons + number of neutrons

So,

Number of neutrons = atomic mass – number of protons

Number of neutrons = 75 – 33 = 42