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Kay [80]
3 years ago
8

In a nuclear reactor, each atom of uranium (of atomic mass 235 u) releases about 200 MeV when it fissions. What is the change in

mass when 2.60 kg of uranium-235 is fissioned?
Physics
1 answer:
Ket [755]3 years ago
7 0

Answer:

0.002372187708 kg

Explanation:

Each atom of Uranium 235 releases 200 MeV = 200×10⁶×1.60218×10⁻¹⁹

= 200×1.60218×10⁻¹³ Joule

Number of atoms in a 2.6 kg sample mass = (2.6/0.235)×6.02214076×10²³

⇒Number of atoms in a 2.6 kg sample mass = 66.627×10²³ atoms

Change in energy = Change in mass / (speed of light)²

ΔE = Δmc²

⇒200×1.60218×10⁻¹³×66.627×10²³ = Δm×(3×10⁸)²

⇒Δm = 200×1.60218×10⁻¹³×66.627×10²³/(3×10⁸)²

⇒Δm = 2372.187708×10⁻⁶ kg

∴Change in mass = 0.002372187708 kg

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<h2>Answer: a bee trying to escape from a closed jar  </h2>

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Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul
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Answer:

t=2.5\times 10^{-14}\ s

Explanation:

We know that charge on electron

q=1.6\times 10^{-19}\ C

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We know that force between two charge given

F=K\dfrac{Q_1Q_2}{r^2}

Now by putting the value

F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}

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m=9.1\times 10^{-31}\ kg

F= m a

a= Acceleration of electron

a= F/m

a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2

a=6.2\times 10^{19} m/s^2

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20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2

t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}

t=2.5\times 10^{-14}\ s

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