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xxTIMURxx [149]
3 years ago
8

In the presence of a nucleus the energy of a y-ray photon can be converted into an electron-positron pair. Calculate the minimum

energy for this to occur and thus the frequency of such a photon. Hint: use E = mc2.]
Physics
1 answer:
Vlada [557]3 years ago
3 0

Answer:

Explanation:

mass of electron in amu = .000549

mass of electron = mass of proton

total mass of electron and proton = 2 x1.098 x 10⁻³ = 2.196 x 10⁻³ amu

1 amu = 931 x 10⁶ eV

2.196 X 10⁻³ amu = 2.196 x 931 x 10³ eV

= 2.044 X 10⁶ eV.

= 2044 X 10³ ev

So minimum energy required = 2044 x 10³ eV.

Wavelength of photon required = 12375/2044 x 10³ A

= 6.054 X 10⁻³A. =6.054 X 10⁻¹³ m

frequency of photon  = 3 x 10⁸/6.054x 10⁻¹³

= .4955 x 10¹⁶Hz.

= 4.955 x 10¹⁵ Hz

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Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

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The orbital speed of satellite A,

v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}......(I)

The orbital speed of satellite B,

v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}

Put the value into the formula

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}

\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}

Hence, The ratio of their orbital speeds are 5:4.

8 0
3 years ago
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B. is the answer.

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