Answer:
These are Diffraction Grating Questions.
Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:
Given as
y = nDλ/w Eqn 1
where
w = width of slit
D = distance to screen
λ = wavelength of light
n = order number
Making x the subject of the formula gives,
w = nDλ/y
Given
y = 0.0149 m
D = 0.555 m
λ = 588 x 10-9 m
and n = 3
w = 6.6x10⁻⁵m
Hence, the width of the slit w, in micrometers (μm) = 66μm
Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen
i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx
Recall Eqn 1, y = nDλ/w
given, D = 27cm = 0.27m
λ = 632 x 10-9 m
w = 0.1mm = 1.0x10⁻⁴m
For the 9th order, n = 9,
y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m
Similarly, for n = 5,
y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m
Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m
Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Answer:
Explanation: El músculo esquelético está formado por fibras musculares, rodeadas de una capa de tejido conjuntivo, denominada endomisio. Las fibras se reúnen en fascículos primarios, que también están rodeados por otra capa de tejido conjuntivo, esta vez, más grueso, denominada perimisio.
A) 0
because all of the forces cancel out, so it is not moving with balanced forces.
Answer:
5.62 m/s
Explanation:
Newton's law of motion can be used to determine the maximum speed of the elevator. In the question, we are given:
Force exerted by the elevator (R) = 1.7 times the weight of the passenger (m*g)
Thus: R = 1.7*m*g
Distance (s) = 2.3 m
Newton's second law of motion: R - m*g = m*a
1.7*m*g - m*g = m*a
a = 0.7*m*g/m = 0.7*g = 0.7*9.8 = 6.86 m/s²
To determine the maximum speed:
Therefore, the elevator maximum speed is equivalent to 5.62 m/s.
