A proton with a speed of 0.9 m/s is directed through a double slit with slit separation 0.4mm. An array of detectors is placed 8m away from the slits. How far off axis is the 2nd minimum in the intensity pattern? Group of answer choices 4.4mm 4.4cm 13.2cm 1.32cm none of the above

vrteclh 2022-08-13 Answered
A proton with a speed of 0.9 m/s is directed through a double slit with slit separation 0.4mm. An array of detectors is placed 8m away from the slits. How far off axis is the 2nd minimum in the intensity pattern?
Group of answer choices
4.4 m m
4.4 c m
13.2 c m
1.32 c m
none of the above
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Answers (1)

Barbara Klein
Answered 2022-08-14 Author has 19 answers
Velocity of the proton is v = 0.9 m / s
Slit separation distance is d = 0.4 m m
Distance of the detector from the slits is D = 8 m
The wavelength of the proton can be calculated using De Broglie's equation
λ = h m v
Here,
λ is the wavelength
h is the Plancks constant
m is the mass
v is the velocity
Substituting the values in the above expression we get,
λ = 6.62 × 10 34 ( 1.66 × 10 27 ) × ( 0.9 ) λ = 4.43 × 10 7 m
Now the distance of the minima in the double slit experiment is given as
Y = ( n + 1 2 ) λ D d
Here,
Y is the distance
n is the integer
Substituting the values in the above expression we get
For second minima, n=1
Y = ( 1 + 1 2 ) ( 4.43 × 10 7 ) × 8 0.4 × 10 3 Y = 0.0132 m = 1.32 c m
Therefore, the distance of the second minima is 1.32 cm.
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