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10 days ago

What wavelength of light contains enough energy in a single photon to ionize a hydrogen atom?

Physics

2 answers:

inna [2.2K]10 days ago

6 0

There might be a quicker and simpler method to approach this, but since I don’t work with these calculations regularly, this is how I do it:

First, I looked up "ionization energy" for Hydrogen on Floogle. This indicates the energy needed to remove one electron from a Hydrogen atom, which is 13.6 eV (electron-volts).

To find the frequency or wavelength of a photon carrying that energy, I must convert the value into Joules.

1 eV = 1.602 x 10⁻¹⁹ Joules (as found on Floogle)

Thus, 13.6 eV = 2.179 x 10⁻¹⁸ Joules

Now we can utilize the familiar formula for photon energy:

Energy = h · (frequency)

or Energy = h · (speed of light/wavelength)

Here, 'h' refers to Max Planck's constant: 6.626 × 10⁻³⁴ m²-kg / s

Excellent! The only unknown in this formula is the wavelength, which we're trying to determine. This should be straightforward now since we know the energy, 'h', and the speed of light.

Wavelength = h · c / energy

Wavelength =

(6.626 x 10⁻³⁴ m²-kg/sec) · (3 x 10⁸ m/s) / (2.179 x 10⁻¹⁸ joule)

Wavelength = 9.117 x 10⁻⁸ meters

This corresponds to 91.1 nanometers.

This wavelength falls short of the visible light spectrum (which ranges from approximately 390 to 780 nm) but is not as brief as I anticipated. I expected it to be classified as X-rays, yet it doesn't reach that short range. X-rays are generally defined within a range of 0.1 to 10 nanometers. The resulting wavelength emerges at the upper end of the Ultra-violet spectrum.

(I am quite pleased with this outcome. I followed the steps I shared with you without any hints. After arriving at my answer, I checked Floogle to see if it matched known values. According to an article on the "Lyman Series," the wavelength for the energy emitted by an electron falling from infinity to the n=1 energy level of Hydrogen is 91.175 nm! I'm thrilled with how close my calculation was, and I am ready to call it a day.)

Sav [2.2K]10 days ago

3 0

The light's wavelength is roughly 9.14 × 10⁻⁸ m

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Further explanation

Max Planck first introduced the concept of quantized energy in electromagnetic radiation, referring to it as photons. The energy quantity is represented as:

$\large {\boxed {E = h \times f}}$

E = Energy of a Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Electromagnetic Wave ( Hz )

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The photoelectric effect occurs when electrons are liberated from a metal surface when exposed to electromagnetic waves with sufficient energy.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

E = Energy of a Photon ( Joule )

m = Electron Mass ( kg )

v = Electron Emission Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Electron Charge ( Coulomb )

V = Stopping Potential ( Volt )

Let us now work through the problem!

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Given:

photon energy = E = 13.6 eV = 2.176 × 10⁻¹⁸ Joule

Unknown:

wavelength of light = λ =?

Solution:

$E = h \times \frac{c}{\lambda}$

$2.176 \times 10^{-18} = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda}$

$2.176 \times 10^{-18} = 1.989 \times 10^{-25} \div \lambda$

$\lambda = (1.989 \times 10^{-25}) \div (2.176 \times 10^{-18})$

$\lambda \approx 9.14 \times 10^{-8} \texttt{ m}$

$\texttt{ }$ Learn more

Photoelectric Effect:
Statements about the Photoelectric Effect:
Rutherford model and Photoelectric Effect:

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics

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Keywords: Quantum, Physics, Photoelectric, Effect, Threshold, Wavelength, Stopping, Potential, Copper, Surface, Ultraviolet, Light

Response: a. The mirrors and eyepiece of a large telescope are designed with spring-loaded components to quickly return to a predetermined position.

Justification:

Adaptive optics refers to a technique employed by various astronomical observatories to compensate in real-time for the atmospheric turbulence that impacts astronomical imaging.

This is executed by integrating advanced deformable mirrors into the telescope's optical pathway, operated by a set of computer-controlled actuators. This allows for obtaining clearer images despite the atmospheric fluctuations that create distortions.

It is crucial to note that this process requires a moderately bright reference star located closely to the object being studied.

However, locating such stars is not always feasible, prompting the use of a strong laser beam directed at the upper atmosphere to create artificial stars.

7 0

4 days ago

An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time

Keith_Richards [2263]

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

3 0

1 month ago

On a hot summer day, you decide to make some iced tea. First, you brew 1.50 LL of hot tea and leave it to steep until it has rea

Yuliya22 [2438]

Explanation:

Please refer to the attachment for the solution.

6 0

10 days ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Ostrovityanka [2208]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

5 0

24 days ago