Question

Consider an atom traveling at 3% of the speed of light. The de Broglie wavelength is found to be 1.42x10^-3 pm. Which element is this?

a) He

b) P

c) F

d) Be

e) Ca

Answer #1

= h/mv

v = 3% of the speed of light = 0.03*3*10^10cm/sec = 9*10^8cm/sec

= 1.42*10^-3 Pm = 1.42*10^-3 *10^-10cm = 1.42*10^-13 cm

h = 6.625*10^-27 erg-sec

m = h/V

= 6.625*10^-27/9*10^8*1.42*10^-13 = 5.18*10^-23g

avogadro' number = 6.023*10^23/mole

atomic mass of element = 5.18*10^-23 *6.023*10^23 = 31.2g/mole >>>>>P

The element is P

b) P>>>>>answer

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light. Its de Broglie wavelength is 3.29x10^{-3}−3 pm. What is
the element's symbol?

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

Calculate the de Broglie wavelength of a Helium atom moving at a
speed of 50 m/s.

Calculate the de Broglie wavelength for a proton with a velocity
3.98% of the speed of light..

Which choice below is correct about the de Broglie
wavelength?
a. The de Broglie wavelength provides a reasonable approximation
of when quantum effects become important.
b. Electrons exhibit a de Broglie wavelength, but protons do
not.
c. The larger the mass of a particle, the longer its de Broglie
wavelength will be.
d. Shorter de Broglie wavelengths correspond to particles with
lower speeds.

1) How will the wavelength of de Broglie electron change if you
increase it
speed 3 times?
2) How much is the wavelength of de Broglie proton different from
the wavelength
where is the Broglie electron if they move at the same speed?
3) Will the wavelength of the particle change if it enters the
potential field?

Find the speed of the following objects given their de Broglie
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b. A proton with a wavelength of 0.2 nm
c. A 200 g baseball with a wavelength of 0.2 nm (in m/s and
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