For a proton (mass = 1.673 x 10–27 kg) moving with a velocity of 2.83 x 104 m/s, what is the de Broglie wavelength (in pm)?
1 answer:
Answer:
The value of de Broglie wavelength is 14.0 pm
Explanation:
Given;
mass of proton, m = 1.673 x 10⁻²⁷ kg
velocity of the proton, v = 2.83 x 10⁴ m/s
De Broglie wavelength is given as;
where;
h is planck's constant = 6.626 x 10⁻³⁴ kgm²/s
m is mass of the proton
v is the velocity of the proton
Therefore, the value of de Broglie wavelength is 14.0 pm
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Answer:
U = 25 J
Explanation:
The energy in a set of charges is given by
U =
in this case we have three charges of equal magnitude
q = q₁ = q₂ = q₃
with the configuration of an equilateral triangle all distances are worth
d = a
U = k ( )
we substitute
15 = k q² (3 / a)
k q² /a = 5
For the second configuration a load is moved to the measured point of the other two
d₁₃ = a
The distance to charge 2 that is at the midpoint of the other two is
d₁₂ = d₂₃ = a / 2
U = k (\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3})
U = k q² ( )
U = (kq² /a) 5
substituting
U = 5 5
U = 25 J
Answer:
The average velocity is 8 unit per sec
Explanation:
Given as :
The Distance x = 20 - 4t² unit
Change in time Δt = ( 2 - 0 ) s = 2 s
Let the velocity = V unit/s
∴ V =
Or, V = - 8t unit/s
Now velocity at t = 0
V1 = - 8 × 0 = 0 unit/s
And velocity at t = 2 sec
V2 = - 8 × 2 = - 16
So, Average velocity = =
= -8
Or, = 8 unit/sec
Hence The average velocity is 8 unit per sec Answer