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JulsSmile [24]
3 years ago
15

A. If an electron of 16 eV had a head-on collision with a Cs atom at rest, what would be the kinetic energy in eV) of the recoil

ing Cs atom? Assume an elastic collision. (Take the atom to be the most abundant isotope of the element. B. In a Frank-Hertz experiment, the first excited state of Hg is obtained at an accelerating voltage measured to be 4.87 V. Based on this measured value, determine the wavelength of the ultraviolet radiation you expect to be emitted.
Physics
1 answer:
Fiesta28 [93]3 years ago
8 0

Answer:

A) 1.96 * 10^-8 J

B) 2.55 * 10^-7 m

Explanation:

A) calculate the kinetic energy in eV of the recoiling Cs atom

we have to apply the principle of energy conservation and momentum

Initial kinetic energy of electron = 16 eV = 16 * (1.6 * 10^-19 ) J

Initial kinetic energy of atom = 0

therefore the final kinetic energy after collision ( E )

= [ 16 * ( 1.6 * 10^-19 ) ] + 0

= 1.96 * 10^-8 J

B) Determine the wavelength of the ultraviolet radiation

accelerating Voltage = 4.87 V

K.E = eV = 1.6 * 10^-19 * 4.87

next we will apply Planck's relationship

\frac{hc}{w} = KE  ---------  ( 1 )

w = wavelength

h = 6.64 * 10^-34 J-S ( Planck's constant )

c = 3 * 10 ^8 m/sec  ( speed of light )

KE = 1.6 * 10^-19 * 4.87

substitute given values into equation 1 above

w ( wavelength ) =  2.55 * 10^-7 m

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Answer:

\frac{d_{1}}{d_{2}}=0.36

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So, the temperature of the first and the second star will be:

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Now the relation between the absolute luminosity and apparent brightness  is given:

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\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

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Answer:

Explanation:

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