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blsea [12.9K]
3 years ago
5

Gamma rays and X–rays are similar because they both have

Physics
1 answer:
VARVARA [1.3K]3 years ago
5 0
<span>High photon energy, high frequency, sub-optical wavelength</span>
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The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec
notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

<u>λ = 1.8 x 10⁻⁷ m = 180 nm</u>

5 0
3 years ago
The wavelength of the sound is 0.750m. What is the frequency?
kifflom [539]

wavelength = speed/frequency


 ==> freq. = speed/wavelength = 342.5/0.75 = 456.67 Hz.

6 0
3 years ago
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denis-greek [22]
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3 0
3 years ago
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both fil
MatroZZZ [7]

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

<u>FOR BULB 1:</u>

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

<u>FOR BULB 2:</u>

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

<u>A₁/A₂ = 0.44</u>

6 0
3 years ago
Think of a famous person and gather information about him/her​
dybincka [34]

Abraham Lincoln

Explanation:

hope it helps you a little

4 0
2 years ago
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