\\lambda = \\dfrac{h}{P} = \\dfrac{h}{\\sqrt{2mE}} \\Rightarrow \\dfrac{h}{\\sqrt{2m(Vq)}} \ herefore For higher m and q ; \\lambda will be smaller.For an ' \\alpha ' particle; both 'm' and 'q' are higherhence lesser is the wavelength.As, (penetrating power) \\propto Energy \\propto \\dfrac{1}{\\lambda} From above; penetrating power of an \\alpha- particle is more than that of a proton.So, the answer is option (B).