The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

(a) 1 mol of He = 4 g = N_A atoms
or 4 g of He $\mathrm{=}\frac{4}{4}\mathrm{mol}=1{\mathrm{N}}_{\mathrm{A}} \mathrm{atom}$
(b) 1 mole of Na = 23 g = N_A atoms $\left[\because \mathrm{no}.\mathrm{of} \mathrm{moles} =\frac{ \mathrm{given} \mathrm{mass} }{ \mathrm{atomic} \mathrm{mass} }\right]$
$\therefore 46 \mathrm{g} \mathrm{of} \mathrm{Na} = \frac{46}{23}\mathrm{mol} = \frac{46}{23}{\mathrm{N}}_{\mathrm{A}} \mathrm{atoms} = 2{\mathrm{N}}_{\mathrm{A}} \mathrm{atoms}$
(c) 1 mole of Ca = 40 g = N_A atoms
$\therefore 0.40\mathrm{g} \mathrm{of} \mathrm{Ca}=\frac{0.40}{40}\mathrm{mol}=\frac{0.40}{40}{\mathrm{N}}_{\mathrm{A}} \mathrm{atoms}$
= 0.01 N_A atoms
(d) 1 mole of He = 4 g = N_A atoms
$\therefore 12 \mathrm{g} \mathrm{of} \mathrm{He}=\frac{12}{4} \mathrm{mol} = \frac{12}{4} {\mathrm{N}}_{\mathrm{A}} \mathrm{atoms} = 3{\mathrm{N}}_{\mathrm{A}} \mathrm{atoms}$