The rubidium isotope 87Rb is a β emitter that has a half-life of 4.9 ✕ 1010 y that decays into 87Sr. It is used to determine the
1 answer:
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Answer:
= 625 nm
Explanation:
We now that for
for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....
d= distance between the slits, λ= wavelength of incident ray
for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.
Given
d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m
applying formula
λ= (d*y)/(D*n)
putting values we get
on solving we get
= 625 nm
The electric current passing through the bulb would be 3.3A
<u>Explanation:</u>
Given:
Electric charge, q = 800C
Time, t = 4 min
= 4 X 60 sec
= 240 sec
Electric current, I = ?
We know,
On substituting the value we get:
Thus, the electric current passing through the bulb would be 3.3A
In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):
Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.
So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:
Therefore the correct option is the second one.