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Marta_Voda [28]
4 years ago
7

A photon of wavelength 3.7 nm Compton scatters from an electron at an angle of 90°. What is the modified wavelength? (Enter your

answer correct to 5 significant figures.)

Physics
2 answers:
viva [34]4 years ago
7 0

Answer:

3.70242 nm

Explanation:

Using Compton effect formula

Δλ = ( h / mec) ( 1 - cosθ)

where h is planck constant = 6.62607 × 10 ⁻³⁴ m²kg/s

me, mass of an electron = 9.11 × 10⁻³¹ kg

c is the speed of light = 3 × 10⁸ m/s

Δλ = 6.62607 × 10 ⁻³⁴ m²kg/s / (9.11 × 10⁻³¹ kg × 3 × 10⁸ m/s ) ( 1 - cos 90°) =   0.242 × 10 ⁻¹¹ m = 2.42 × 10⁻¹² m  = 0.00242 nm

modified wavelength = 3.7 nm + 0.00242 nm = 3.70242 nm

denis-greek [22]4 years ago
7 0

Answer: 3.7024nm

Explanation:

in the attachment

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Fermi energy of conduction electrons in silver is 0.548 J. Calculate the number of such electrons in unit cm3
ICE Princess25 [194]

Answer:

n=1.86*10^{-30}m^3

Explanation:

From the question we are told that:

Fermi energy of conduction electrons E_f=0.548J

Generally the equation for Fermi energy is mathematically given by

 E_f=\frac{3}{\pi}^2/3*\frac{h^2}{8m}*{n^{2/3}}

 E_f=\frac{3}{\pi}^2/3*\frac{h^2}{8m}*{n^{2/3}}

 {n^{2/3}}=\frac{E_f}{\frac{3}{\pi}^2/3*\frac{h^2}{8m}}

Where

h= Planck's constant

 {n^{2/3}}=\frac{E_f}{\frac{3}{\pi}^2/3*\frac{h^2}{8m}}

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 n=(1.51*10^{-20})^{3/2}

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3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
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pickupchik [31]
Her average speed per hour, was 42.5 miles per hour
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3 years ago
A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du
Lana71 [14]

Answer:

a_E=\dfrac{Gm}{r^2}

Explanation:

M_E = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}

The acceleration that the Earth will feel is a_E=\dfrac{Gm}{r^2}

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Juli2301 [7.4K]

Answer:

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Explanation:

Given data

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In order to determine the time required to melt the ice, we can use the following expression.

P = Q/t

t = Q / P = 33,200 J/ 125 W = 266 s

It takes 266 seconds to melt the ice.

5 0
3 years ago
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