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Black_prince [1.1K]
2 years ago
13

X-ray diffraction is used to study the structure of crystallized proteins, nucleic acids, and other biological macromolecules. I

t was actually the technique used to study the structure of DNA in the early 1950s. X-rays of wavelength 0.20 nm are to used to study the structure of a protein. One of the maxima in intensity is located at 0.8 rad from the crystal planes responsible for this maximum.
PART A
Which wavelength of x-ray listed below would give you a maximum at the same location?
a. .26nm
b. .43nm
c. .34nm
d. .60nm

PART B
If this is the first maximum for the X-rays of wavelength 0.17 nm , what is the crystal plane separation of the protein that is responsible?
a. .12nm
b. .24nm
c. .17nm
d. .11nm

PART C
Why can't visible light be used to study the structure of proteins such as this?
a. Visible light reflects off of the surface of proteins.
b. Visible light does not have enough energy.
c. The wavelengths of visible light are too long compared to the atomic spacing.
d. The wavelengths of visible light are too short compared to the atomic spacing.
Physics
1 answer:
stepladder [879]2 years ago
7 0

Answer:

PART A: option b. .43nm

PART B: option d. 0.11nm

PART C: option c. The wavelengths of visible light are too long compared to the atomic spacing.

Explanation:

Given data

Wavelength λ = 0.20 nm

Angle θ = 0.8 rad

(a)

wavelength of x-ray to give maximum at the same location

λ₂ = m λ

Here, m = 2 is the interference fringe order.

Substitute the values in the above equation.

λ₂ = 2 × 0.2

    = 0.4 nm

Hence, the wavelength of x-ray to give maximum at the same location is 0.4nm

(b)

The crystal plane separation is equal to d

The value of θ is equal to 0.8 rad.

Convert rad into degree as follows:

0.8 rad = \frac{180^0}{\pi rad} (\frac{0.8rad}{1}) = 144°/π = 45.86°

Solve for d, using equation (1) as follows:

2dsinθ = mλ

d = mλ / 2sinθ

d = (1) 0.17 / 2Sin45.86°

d = 0.17 / 1.9065

d = 0.089 nm

(c)

The visible light can not be used to study the structure of proteins because of the high wavelength of the visible light.

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Answer:

33.2 m

Explanation:

For the first object:

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t₀ = 0 s

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For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

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A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
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Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

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