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2 years ago

10

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]2 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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So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

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$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

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$h=4.136(10)^{-15}eV.s$ is the Planck constant

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Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

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$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

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$E_{2}=3.011eV$ (13)

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$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

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