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Pani-rosa [81]
3 years ago
5

Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul

ting in electrons being ejected from the atoms. N.B. The energy of these photons is much smaller than the rest energy of the electron, so you don't need relativistic expressions. Assume that the photon is destroyed in this process, leaving only the electron moving away from a stationary hydrogen ion.
a. With the gas in thermal equilibrium at room temperature, what would be the maximum kinetic energy of ejected electrons?
b. When an intense light source of the same wavelength is shone on the gas sample, a small number of electrons with kinetic energy as much as 10.2 eV greater than you calculated in part a. Explain how this can be.
Physics
1 answer:
timofeeve [1]3 years ago
8 0

Answer:

a)   K_{max} = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

      K_{max}= h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

     c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

     Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

     E = 2,486 10⁻¹⁸ J

Let's reduce to eV

     E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

     E = 15.5 eV

Now we can calculate the kinetic energy

     K_{max}= h c / f - fi

      K_{max} = 15.5 -13.6

     K_{max} = 1.9 eV

b)     Extra energy = 10.2 eV

The total kinetic energy of electrons is

       Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

      ΔE = E photon - Total kinetic energy electron

      ΔE = 15.5 - 12.1

      ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

     E_{n} = 13.606 / n2

     n = √ 13.606 / En

     n = √ (13606 / 3.4)

     n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

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Mila [183]

Answer:

5.731\times 10^{-5}\ m/s

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = 1.6\times 10^{-19}\ C

n = Conduction electron density in copper = 8.49\times 10^{28}\ electrons/m^3

v_d = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s

The drift speed of the electrons is 5.731\times 10^{-5}\ m/s

v_d=\dfrac{I}{neA}

From the equation we can see the following

v_d\propto \dfrac{1}{n}

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0
3 years ago
An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

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8 0
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A skydiver has jumped out of a plane and is falling faster and faster. what forces are present in this situation
kompoz [17]
Gravity and air resistance 

i took the test and got 100%
5 0
3 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
zzz [600]

Answer:

1000 Hz

Explanation:

<em>The frequency would be 1000 Hz.</em>

The frequency, wavelength, and speed of a wave are related by the equation:

<em>v = fλ ..................(1)</em>

where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.

Making f the subject of the formula:

<em>f = v/λ.........................(2)</em>

Also, speed (v) = distance/time.

From the question, distance = 900 m, time = 3.0 s

Hence, v = 900/3.0 = 300 m/s

Substitute v = 300 and λ = 0.3  into equation (2):

f = 300/0.3 = 1000 Hz

6 0
3 years ago
You observe three carts moving to the left. Cart A moves to the left at nearly constant speed. Cart B moves to the left, gradual
Lady bird [3.3K]

Answer:cart B

Explanation:

For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity

thus there is no net force

For cart B there is change in velocity in the left direction , so there is net acceleration towards left

Force=mass\times acceleration

so there is net force in the left direction

For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion                

6 0
3 years ago
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