The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
In our problem, we know
and 
, and using the relationship mentioned above, we can find the mass m of the particle:
In our problem, we know
What is the wavelength of the matter wave associated with a proton moving at 377377 m/s? wavelength of proton matter wave: mm Wh
1 answer:
1)
The wavelength of the matter wave (also called de Broglie wavelength) of an object is given by
where
is the Planck constant
m is the mass of the object
v is its velocity
For a proton, we have:
and the velocity of this proton is
So, its de Broglie's wavelength is:
2)
We can use again the same equation:
where in this case we have:
m = 159 kg is the mass of the astronaut + spacesuit
v = 377 m/s is the velocity of the astronaut
Substituting into the equation,
3)
Similarly, we can use the same equation:
where in this case we have: is the Earth's mass
is the velocity of the Earth around the Sun
Substituting,
You might be interested in
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 / - 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.