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iris [78.8K]
3 years ago
14

How many 100W light bulbs could be powered for one year by the direct conversion of 1g of matter into energy?

Physics
1 answer:
stellarik [79]3 years ago
5 0

Answer:

No. of 100 W bulbs, n = 28,539 bulbs

Given:

Power of a single bulb = 100 W

Time period, T = 1 yr = 365\times 24\times 60\times 60 = 31,536,000 s

mass of matter, m = 1 g = 1\times 10^{-3}]

Solution:

According to Eintein's mass-energy equivalence:

E = mc^{2}

E = 1\times 10^{-3}\times (3\times 10^{8})^{2}

E =  9\times 10^{13} J

Power of a single bulb, P =  \frac{E}{T}

                                       P =  \frac{9\times 10^{13} }{31,536,000}

                                       P = 28,53,881 W

No. of 100 W bulbs, n = \frac{P}{power of one bulb}

                                 n = \frac{28,53,881}{100}

                                 n = 28,539

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
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Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

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Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

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According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

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where,

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For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

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For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

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