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3 years ago

Why do X-Ray technicians stand behind a shield when giving X-Rays to patients?

2 answers:

7 0

Lol I just learned this, here's the short answer:
Large doses of X-Rays are known to cause cell damage and cancer, so
<span>X-Ray technicians have to stand behind the shield. Here's the lengthy answer:
</span> This is because the patient would only be exposed to the small dose of radiation once, which is hardly harmful. However, radiographers deal with this all day every day, which could mean that the small doses could add up. This is why they stand behind a lead screen; the small exposure to the X-Rays aren't harmful alone but frequent exposure could begin to damage your cells and lead to potential genetic mutation. If a patient is often in radiography (a long-stay patient for example), they begin to wear lead aprons whilst being X-rayed in order to prevent the risk of harm. Hope I answered correctly.

Mashcka [7]3 years ago

7 0

Answer:

because large doses of X-Rays are known to cause cell damage and cancers

Explanation:^^

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An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t

SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f) in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0

3 years ago

How do you measure the wavelength of wave?

Tasya [4]

Explanation:

by finding the distance between two successful crest

6 0

3 years ago

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Realiza la siguiente conversión de unidades: 340 N a Kgf

OleMash [197]

Answer:

The answer is 34.67 kilograms force

3 0

3 years ago

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng

klemol [59]

Answer:

Speed:

$2.01x10^{8}m/s$

Wavelength:

$4.24x10^{-7}m$

Frequency:

$4.74x10^{14}Hz$

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

$n = \frac{c}{v}$ (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

$v = \frac{c}{n}$

$v = \frac{3x10^{8}m/s}{1.490}$

$v = 2.01x10^{8}m/s$

Hence, the speed of the laser has a value of $2.01x10^{8}m/s$

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.

To determine the frequency it can be used the following equation

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength

Then, $\nu$ wil be isolated from equation 2.

$\nu = \frac{c}{\lambda}$ (3)

Before using equation 3 it is necessary to express $\lamba$ in units of meters.

$\lambda = 632.8nm . \frac{1m}{1x10^{9}nm}$ ⇒ $6.328x10^{-7}m$

$\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}$

$\nu = 4.74x10^{14}s^{-1}$

$\nu = 4.74x10^{14}Hz$

Hence, the frequency of the laser has a value of $4.74x10^{14}Hz$

Wavelength:

To determine the wavelength it can be used:

$v = \nu \cdot \lambda$

$\lambda = \frac{v}{\nu}$

Where v is the speed of the laser through the polystyrene.

$\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}$

$\lambda = 4.24x10^{-7}m$

Hence, the wavelength of the laser has a value of $4.24x10^{-7}m$

3 0

3 years ago

A boy rides a sled down a steep, snow.covered hill. Which of

Travka [436]

Answer:

a. Potential energy decreases and Kinetic energy increases

Explanation:

Because as he comes down due to its steepness the speed of the boy or you can say his KE increases and since he comes from a high position (hill) to the lower ground his potential energy decreases simultaneously

4 0

3 years ago

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