Answer:
39.40 MeV
Explanation:
<u>Determine the minimum possible Kinetic energy </u>
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
<em>Steel: 11.0 – 12.5</em>
<em>T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶ ̶T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶</em>
Thanks,
<em>Deku ❤</em>
Answer:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."
Explanation:
The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.
D sin θ = m λ
if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."
This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.
Hope this helps!
