In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
(2)
If we use the formula to rewrite
, eq.(2) becomes
Which we can solve to find a, the width of the slit:
Answer: 1.39 s
Explanation:
We can solve this problem with the following equations:
(1)
(2)
Where:
is the length the steel wire streches (taking into account 1mm=0.001 m)
is the length of the steel wire before being streched
is the force due gravity (the weight) acting on the pendulum with mass 
is the transversal area of the wire
is the Young modulus for steel
is the period of the pendulum
is the acceleration due gravity
Knowing this, let's begin by finding :
(3)
Where 
is the diameter of the wire
(4)
(5)
Knowing this area we can isolate from (1):
(6)
And substitute in (2):
(7)
(8)
Finally:
Answer:
a) 0.0288 grams
b) 
Explanation:
Given that:
A typical human body contains about 3.0 grams of Potassium per kilogram of body mass
The abundance for the three isotopes are:
Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.
a)
Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.
However, the amount of potassium that is present in such person is :
0.012% × 240 grams
= 0.012/100 × 240 grams
= 0.0288 grams
b)
the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:
First the Dose in (Gy) = 
= 
= 
Effective dose (Sv) = RBE × Dose in Gy
Effective dose (Sv) = 
Effective dose (Sv) =
