Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is
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Answer:
The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.
Explanation:
Formula used for the radius of the orbit will be,
where,
= energy of
orbit
n = number of orbit
Z = atomic number
Here: Z = 1 (hydrogen atom)
Energy of the first orbit in H atom .
Energy of the second orbit in H atom .
Energy of the third orbit in H atom .
Energy of the fifth orbit in H atom .
Energy of the sixth orbit in H atom .
Energy of the seventh orbit in H atom .
During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :
1) n = 2 → n = 3
2) n= 5 → n = 6
Energy absorbed when: n = 2 → n = 3
Energy absorbed when: n = 5 → n = 6
1.89 eV > 0.166 eV
E> E'
So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.
Answer:
The final image relative to the converging lens is 34 cm.
Explanation:
Given that,
Focal length of diverging lens = -12.0 cm
Focal length of converging lens = 34.0 cm
Height of object = 2.0 cm
Distance of object = 12 cm
Because object at focal point
We need to calculate the image distance of diverging lens
Using formula of lens
The rays are parallel to the principle axis after passing from the diverging lens.
We need to calculate the image distance of converging lens
Now, object distance is ∞
Using formula of lens
The image distance is 34 cm right to the converging lens.
Hence, The final image relative to the converging lens is 34 cm.
The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7º
given parameters
- the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
- the movement times t = 1.0s, 2.0s and 3.0 s
to find
a) position
b) acceleration
c) launch angle
Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration
b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.
v_y = v_{oy} - g t
where v and v({oy} are the velocities of the body, g the acceleration of the planet's gravity and t the time
0 = v_{oy} - gt
g = v_{oy} / t
from the graph we observe that the highest point occurs for t = 2.0 s
g = 1.2 / 2.0
g = 0.6 m / s²
a) The position is requested for several times
X axis
in this axis there is no acceleration so we can use the uniform motion relationships
vₓ = x / t
x = vₓ t
where x is the position, vx is the velocity and t is the time
we calculate for the time
t = 0.0 s
x₀ = 0
t = 2.0 s
x₂ = 1.8 2
x₂ = 3.6 m
t = 3.0 s
x₃ = 1.8 3
x₃ = 5.4 m
Y axis
In this axis there is the acceleration of the planet, let us use for the position the relation
y = v_{oy} t - ½ g t²
t = 0.0 s
y₀ = 0
y₀ = 0 m
t = 2.0 s
y₂ = 1.2 2 - ½ 0.6 2²
y₂ = 1.2 m
t = 3.0 s
y₃ = 1.2 3 - ½ 0.6 3²
y₃ = 0.9 m
c) the launch angle use the trigonometry relation
tan θ =
θ = tan⁻¹
θ = tan⁻¹
θ = 33.7º
measured counterclockwise from the positive side of the x-axis
With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7ºto)
learn more about projectile launch here: