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Valentin [98]
3 years ago
7

The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.What power is absorbed by the bulb?

Physics
1 answer:
Verizon [17]3 years ago
3 0

Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

Power = Voltage × Current

Given:

Voltage = 3 V

Current = 200 mA = 0.200 A

Conversion used : (1 mA = 0.001 A)

Now put all the given values in the above formula, we get:

Power = Voltage × Current

Power = 3V × 0.200 A

Power = 0.600 W

Thus, the power absorbed by the bulb is, 0.600 W

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Explanation:

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\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

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Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

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From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

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P=h/\lambda and putting it in the above equation we get,

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Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

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