(a)T2248i(b)S3479e(c)B511Draw atomic depictions similar to those in Problem 2.43 for $\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}{}_{\mathbf{22}}{}^{\mathbf{48}}\mathit{T}\mathit{i}\mathbf{}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}{}_{\mathbf{34}}{}^{\mathbf{79}}\mathit{S}\mathit{e}\mathbf{}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}{}_{\mathbf{5}}{}^{\mathbf{11}}\mathit{B}$

${}_{22}{}^{48}Ti$

Given,

Atomic number = 22

Atomic mass = 48

Number of protons = Atomic number = 22

Number of electrons = 22

We know,

Atomic mass = number of protons + number of neutrons

So,

Number of neutrons = atomic mass – number of protons

Number of neutrons = 48 – 22 = 26

${}_{34}{}^{79}Se$

Given,

Atomic number = 34

Atomic mass = 79

Number of protons = Atomic number = 34

Number of electrons = 34

We know,

Atomic mass = number of protons + number of neutrons

So,

Number of neutrons = atomic mass – number of protons

Number of neutrons = 79 – 34 = 45

${}_5{}^{11}B$

Given,

Atomic number = 5

Atomic mass = 11

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 = 11 – 5 = 6