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 3x10^{8} m/s

Hence 1.5 % of the speed of light = 3x10^{8} m/s x (1.5
/ 100) = 4.5x10^{6} m/s

Hence the speed of the atom, V = 4.5x10^{6} 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.m^{2}.s^{-2}) / (8.6 x
10^{-16} m x4.5x10^{6} 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
6.023x10^{23} = 103 g

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

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