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Alex787 [66]
3 years ago
15

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

t the emissivity eee is equal to 1 for these surfaces.
Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.
Physics
1 answer:
belka [17]3 years ago
8 0

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

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When light of wavelength 160 nm falls on a gold surface, electrons having a maximum kinetic energy of 2. 66 ev are emitted. Find
enyata [817]

When the light of wavelength is falling on gold surface, the electrons begin to exchange energies.

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

<h3>What is work function?</h3>

The energy needed for a particle to escape and break through the surface.

The kinetic energy of the light emitted is 2.66 eV and wavelength of the light is 160 nm = 160 × 10⁻⁹ m.

a) The work function of the gold for given maximum kinetic energy is

Φ = hc / λ  - K.Emax

Substituting 6.626 × 10⁻³⁴ J.s for h, 3 × 10⁸ m/s for c and 2.66 eV for K.Emax, work function will be

Φ =8.16 × 10⁻¹⁹ J

1 eV = 1.6 × 10⁻¹⁹

The work function in eV is Φ =5.097 eV.

b) The cutoff wavelength is related to work function as

λ₀ = hc / Φ

Substitute the corresponding values into the equation, we get the cut off wavelength

λ₀ = 243.71 nm

c) The frequency corresponding to the cut-off wavelength is

ν₀ = c / λ₀

Substitute the corresponding values into the equation, we get the frequency,

ν₀  =1.231 × 10¹⁵ Hz

Therefore, the values for the following are

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

Learn more about wave function.

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3 years ago
4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav
Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

                                                                         = 4.5 x 10⁹ Km

                                                                          = 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

                                           <em>V = d / t    m/s</em>

∴                                           t = d / V

                                               = 4.5 x 10¹¹ m / 3 x 10⁸ m/s

                                               = 15,000 s

                                               = 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

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