Question

A single atom is accelerated to a velocity of 1.5% the speed of light. At this speed the atom has a de Broglie wavelength of 8.6×10-16 m. What is the atom in question?

(a) Helium (He)

(b) Lutetium (Lu)

(c) Rhodium (Rh)

(d) Hydrogen(H)

(e) An atom cannot have a wavelength.

**solutions say the anser is (C) but i dont know how they got that

Answer 1

The speed of light is 3x10⁸ m/s

Hence 1.5 % of the speed of light = 3x10⁸ m/s x (1.5 / 100) = 4.5x10⁶ m/s

Hence the speed of the atom, V = 4.5x10⁶ m/s.

The corresponding de Broglie wavelength of the atom,   = 8.6 x 10^-16 m

According to wave nature of matter, the wave length associated with a certain mass is inversely ptoportional to the momentum. Hence

(lambda) is inversely proportional to P (= mxv)

=> (lambda) = h / mxv

=> m = h /  (lambda) x v = (6.63x10^-34 Kg.m².s^-2) / (8.6 x 10^-16 m x4.5x10⁶ m/s) = 1.713x10^-25 Kg

=> m =  1.713x10^-25 Kg =  1.713x10^-25 Kg x (1000 g / 1 Kg) = 1.713x10^-22 g

Hence m = 1.713x10^-22 g is the mass of a single atom. Now we can get the atomic mass of the atom by multiplying the mass of single atom by Avogadro's number.

Hence the atomic mass of the unknown atom

= mxN_A =  1.713x10^-22 g x 6.023x10²³  = 103 g

103 g is the atomic mass of Rhodium (Rh). Hence the atom is Rhodium (Rh) (answer)