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hichkok12 [17]
3 years ago
11

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

e between these wavelengths?
Physics
1 answer:
Strike441 [17]3 years ago
6 0

The solution would be like this for this specific problem:

 

Given:  

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

 

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m <span>

Whenas n = 1, maxima (grating equation) sinθ = λ/d 
Grant distance of each maxima from centre = y .. 
<span>As sinθ ≈ y/D  y/D = λ/d λ = yd / D </span>

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D 
∆λ = (d/D) [y2 -y1] 

<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>

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Answer:

The longest wavelength of radiation is 241nm and it lies in ultraviolet region.

Explanation:

The minimum energy required to break o-o bond is 495kJ/mole.

The photon does not have mass and the energy of the single photon depends entirely on the wavelength and is given by

e =hc/λ

where, h is the Planck constant,

            c is the speed of light

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From the Planck formula we can understand that energy of the photon is quantized.

E = e.Nₐ

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Lindsey started biking to the park traveling 15 mph, after some time the bike got a flat so Lindsey walked the rest of the way,
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And the time she spent walking that distance = (5 - t) hours

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Substitute for X in (eqn 2)

53 - 15t = 20 - 4t

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x = 15t = 15 × 3 = 45 miles.

(53 - x) = 53 - 45 = 8 miles

(5 - t) = 5 - 3 = 2 hours

So, it becomes evident that Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

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