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adell [148]
3 years ago
13

What potential increase vac must an electron be accelerated through if the most energetic photon it can emit will scatter off of

a stationary electron at an angle ϕ=60∘ with wavelength 8.4×10−12m?express your answer in kilovolts to two significant digits?
Physics
1 answer:
nikitadnepr [17]3 years ago
7 0
Referring to Compton scattering 
Δλ = h/m₀c (I- cos Ф)
λ' =λ = (0,0242×10⁻¹⁰) (1- cos 60°)
λ= λ' -(0.0242 × 10⁻¹⁰) (1- cos 60°)

7.19 ˣ 10⁻¹²m

The increased potential is given by 
Vₐc = hc/eλ = 6.625 × 10 ⁻³⁴ J,s) ( 3× 10⁸ m/s ( 1.6 ˣ 10 ⁻¹⁰C)
(7.19 ˣ 10⁻¹²m)

173kV.
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3 years ago
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An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value
Dimas [21]

Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

R = Rydberg constant, R=1.097\times 10^7\ m^{-1}

\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

Solving above equation we get the value of final n is,

n_f=2.04

or

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A diverging lens of focal length 18.0m is used to view a shark that is 90.0m away from the lens. If the image formed is 1.0m lon
Reil [10]

Answer:

i. + 22.5 m ii. 4.0 m

Explanation:

i. Image distance

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v = 90/4

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ii Length of Shark

Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.

m = image distance/object distance

= v/u

= +22.5/+90

= 0.25

0.25 = image height/object height

So,

object height = image height/0.25

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= 4.0 m

So, the length of the shark is 4.0 m

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In a motion diagram, the size of the object must be less than the distance moved.
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Is there  more to this question
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