If you know the amount of the unbalanced force acting upon an object and the mass of the object, using Newton's 2nd Law what cou
2 answers:
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Answer:
a)
For this case we know the following values:
So then if we replace we got:
b)
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
c)
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
Answer:
Proton: v=0.689 m/s
Neutron: v=0.688 m/s
Electron: v=1265.078 m/s
Alpha particle: v=0.173 m/s
Explanation:
De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:
λ=
h is the Planck constant: 6.626×10⁻³⁴
We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:
λ=
v=h÷(mλ)
<u>Proton:</u>
m=1.673×10⁻²⁴ g · =1.673×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)
v=0.689 m/s
<u>Neutron:</u>
m=1.675×10⁻²⁴ g · =1.675×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)
v=0.688 m/s
<u>Electron:</u>
m= 9.109×10⁻²⁸ g · =9.109×10⁻³¹ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(9.109×10⁻³¹ kg×575×10⁻⁹m)
v=1265.078 m/s
<u>Alpha particle:</u>
m=6.645×10⁻²⁴ g · =6.645×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)
v=0.173 m/s