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kakasveta [241]
2 years ago
10

If the maximum energy given to an electron during compton scattering is 30 kev, what is the wavelength of the incident photon?

Physics
1 answer:
hammer [34]2 years ago
6 0

The wavelength of the incident photon is 1.19x10^{-2} nm.

What is wavelength?

The wavelength, or the distance over which the shape of a periodic wave repeats, is the spatial period in physics. It is a property of both traveling waves and standing waves, as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two neighboring crests, troughs, or zero crossings. The Greek letter lambda is frequently used to denote wavelength. The term wavelength is also sometimes used to describe modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

The relationship between wavelength and frequency is inverse, assuming a sinusoidal wave flowing at a constant speed.

Calculations:

The energy loss Δλ=h/m_{e} c(1-cos∅)

Conservation of momentum gives,

h_{f} =P_{e} c-hf'\\hf+hf'=P_{e} c\\hf+hf'=\sqrt{(m_{e} c^{2} } +k^{2} )-me^{2} c^{4} \\\\=\sqrt {(511+30)^{2} -(511)^{2} } \\=178keV

hf=hf'+k=hf-hf'=30keV\\

2hf=178+30\\2hf=208\\hf=104keV\\

Wavelength(λ)=\frac{hc}{104keV}=\frac{1.24keV.nm}{104keV}

Wavelength(λ)=1.19x10^{-2} nm

To learn more about wavelength , visit:

brainly.com/question/12924624

#SPJ4

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Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is
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Answer:

d \frac{x}{l} = m×  λ⇒ d = λ ×m×l / x

= 630×10^{-9} m × 3×3m/ 45×10^{-3} m

= 1.26×10^{-4}m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

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3 years ago
A star has an absolute magnitude of 4 and a surface temperature of 5,000 degrees C. According to the HR diagram, list the type o
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A common pickup for an electric guitar consists of a coil of wire around a small permanent magnet, as described in Figure 25.5.
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6 0
3 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
3 years ago
Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f
Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

\lambda=380 nm =3.8\cdot 10^{-7}m (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

Substituting,

E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J

And keeping in mind that

1 eV = 1.6\cdot 10^{-19}J

This energy converted into electronvolts is

E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0
3 years ago
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