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3 years ago

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A positron and an electron annihilate each other upon colliding, thereby producing energy in the form of two gamma rays. Assumin

g that both gamma rays have the same energy (since both particles have the same mass), calculate the wavelength of the electromagnetic radiation used in pm. (1 pm = 10⁻¹² m; mass of electron (amu) = 0.000549)

1 answer:

kolbaska11 [484]3 years ago

6 0

Answer:

2.42631E-13m

Explanation:

First we find the mass defect

Which is m= 0-2(9.10939E-33kg)

= - 1.82188E-30kg

Now find the energy

S

E= mc²=( -18.82188E-30)(2.999792E8)²

= 1.63742E-13J

Thus energy per photon will be

1.63742E-13J/2= 8.18710E-14J

So wavelength is given as

Lambda= hc/E

= (6.62608E-34)(2.997E8)/8.18710J

= 2.42631E-13m

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Answer:

a. <u>position function of the coin:</u>

$s=-16t^2+1344$

<u>Now the velocity function:</u>

$v=-32t$

b. $v_{avg}=-112\ m.s^{-1}$

c. $v_3=-96\ m.s^{-1}$ & $v_4=-128\ m.s^{-1}$

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Explanation:

Given:

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Given position function associated with free falling objects:

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here:

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$v_0=$ initial velocity

$t=$ time of observation

a)

<u>position function of the coin:</u>

$s=-16t^2+1344$

∵the object is dropped it was initially at rest

<u>Now the velocity function:</u>

$v=\frac{d}{dt} s$

$v=-32t$

b)

we know average velocity is given as:

$\rm v_{avg}=\frac{total\ displacement}{total\ time}$

Displacement in the given interval:

$s_{_{3-4}}=s_4-s_3$

$s_{_{3-4}}=(-16\times 4^2+1344)-(-16\times 3^2+1344)$

$s_{_{3-4}}=-112\ ft$

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$v_{avg}=\frac{-112}{4-3}$

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c)

Instantaneous velocity at t = 3 s:

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$v_3=-96\ m.s^{-1}$

Instantaneous velocity at t = 4 s:

$v_4=-32\times 4$

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d)

At ground we have s=0:

Put this in position function:

$0=-16t^2+1344$

$t=9.165\ s$

e)

Velocity of the coin at impact:

$v_f=-32\times 9.165$

$v_f=293.285\ m.s^{-1}$

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Answer:

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