When ultraviolet light with a wavelength of 400. 0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1. 10 ev. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300. 0 nm falls on the same surface?.

Answers

Answer 1

The maximum kinetic energy of the photoelectron K0 when light with a wavelength of 310 nm falls on the same surface is 2.135 eV.

You need to find a working function to solve this problem. Quantum Theory Regarding the Photoelectric Effect of the work function (eV), the work function describes the minimum energy required for electrons to continue to stick to the metal.

Planck's constant is also 4.14 x 10⁻¹⁵ eV.

Max KE = 1.10 eV

lambda = 400 nm = 400 x 10⁻⁹ m

First, find the frequency.

f = c ÷ λ

f = (3 x 10⁸) ÷ (400 x 10⁻⁹ m)

f = 7.5 x 10¹⁴

Max KE = hf - work function

1.10 eV = (4.14 x 10⁻¹⁵eV) (7.5 x 10¹⁴) - function

work function = 2.005 eV

Now find the frequency for the second wavelength.

f = c /λ

f = (3 x 10⁸) ÷ (300 x 10⁻⁹)

f = 1 x 10¹⁵

Now plug this into your equation with the frequency and work function you found in the previous problem.

KEmax = hf - work function

KEmax = (4.14 x 10⁻¹⁵ eV)(1 x 10¹⁵) - 2.005 eV

KEmax = 2.135 eV

So, the maximum kinetic energy is 2.005 eV

The complete answer:

When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 eV. What is the maximum kinetic energy K0 of the photoelectrons when the light of wavelength 310 nm falls on the same surface? Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.

Learn more about the work function at https://brainly.com/question/19427469.

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Exact answer from Plato. Let's graduate together lol ; )

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the frequency of a wave in a stretched string depends on the length L, tension T and linear density m with dimension ML-1 . deduce the formula for f in terms of L , T and M using dimensional analysis​

Answers

Answer:

[tex]f = \frac{1 }{L}\sqrt{\frac{T}{m} }[/tex]

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Let the frequency ,

[tex]f = L^{x}T^{y}m^{z}[/tex]

Now the unit of frequency is hertz = s⁻¹ = T⁻¹ where T is time, tension T = kgm/s² = MLT⁻¹ and linear density m = kg/m = ML⁻¹.

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[tex]T^{-1} = L^{x}[MLT^{-2} ]^{y}[ML^{-1} ]^{z}\\[/tex]

collecting the like bases, we have

[tex]T^{-1} = [L^{x + y -z}][M^{y + z} ][T^{-2y} ] \\L^{0} M^{0} T^{-1} = [L^{x + y -z}][M^{y + z} ][T^{-2y} ][/tex]

Equating powers on both sides, we have

x + y - z = 0  (1)

y + z = 0       (2)

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From (3), y = 1/2

From (2), z = -y = -1/2

Substituting y and z into (1), we have

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So,

[tex]f = L^{x}T^{y}m^{z}\\f = L^{-1}T^{\frac{1}{2} }m^{\frac{-1}{2} }[/tex]

[tex]f = \frac{1 }{L}\sqrt{\frac{T}{m} }[/tex]

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