Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
The smallest unit that makes up matter is an atom!
C. The downward component of the projectile's velocity continually increases
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion (with constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction
Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

where
u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)
downward is the acceleration of gravity
t is the time
So the equation becomes

This means that
C. The downward component of the projectile's velocity continually increases
Because every second, it increases by
in the downward direction.
Learn more about projectile motion:
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That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.