Question

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 . part a what is the maximum kinetic energy k0 of the photoelectrons when light of wavelength 270 nm falls on the same surface?

Answer 1

In the photoelectric effect, the energy carried by the incoming photon is used partially to extract the photoelectron from the material  and the rest is converted into kinetic energy of the electron:
$hf = \phi + K_{max}$ (1)
where
h is the Planck constant
f is the photon frequency
$\phi$ is the work function of the material (the energy needed to extract the photoelectron)
$K_{max}$ is the maximum kinetic energy of the photoelectron

The ultraviolet light has a wavelength of $\lambda=400 nm=400 \cdot 10^{-9} m$, so its frequency is
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{400 \cdot 10^{-9} m}=7.5 \cdot 10^{14}Hz$
And the energy of each photon of this light is
$E=hf=(6.6 \cdot 10^{-34}Js)(7.5 \cdot 10^{14}Hz)=4.95 \cdot 10^{-19}J$

The maximum kinetic energy of the emitted photoelectrons is (converting into Joules)
$K_{max} = 1.10 eV \cdot 1.6 \cdot 10^{-19} J/eV =1.76 \cdot 10^{-19}J$

And so by using equation (1) we can find the work function of the material:
$\phi = hf - K_{max} = 4.95 \cdot 10^{-19}J - 1.76 \cdot 10^{-19} J =3.19 \cdot 10^{-19}J$

Now we have instead light wavelength $\lambda=270 nm= 270 \cdot 10^{-9}m$ hitting the same surface. The frequency of this light is
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{270 \cdot 10^{-9}m}=1.11 \cdot 10^{15} Hz$
and the energy of each photon of this light is
$E=hf=(6.6 \cdot 10^{-34}Js)(1.11 \cdot 10^{15}Hz)=7.33 \cdot 10^{-19} J$

And so we can calculate the new maximum kinetic energy of the photoelectrons by using (1) and the work function we found previously:
$K_{max} = hf-\phi = 7.33 \cdot 10^{-19}J - 3.19 \cdot 10^{-19}J = 4.14 \cdot 10^{-19}J$

And if we want to convert it into electronvolts,
$K_{max} = \frac{4.14 \cdot 10^{-19}J}{1.6 \cdot 10^{-19} J/eV}=2.59 eV$ 

The maximum kinetic energy of the photoelectrons has increased because the light hitting the surface is now more energetic, so it can transfer more energy to the electrons.

Answer 2

The maximum kinetic energy is about 2.59 eV

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Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

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

E = Energi of A Photon ( Joule )

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

f = Frequency of Eletromagnetic Wave ( Hz )

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The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

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

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

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem !

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Given:

wavelength of first light = λ₁ = 400 nm = 4 × 10⁻⁷ m

the first maximum kinetic energy = Ek₁ = 1.10 eV = 1.76 × 10⁻¹⁹ J

wavelength of second light = λ₂ = 270 nm = 2.7 × 10⁻⁷ m

Asked:

the second maximum kinetic energy = Ek₂ = ?

Solution:

$E_1 = Ek_1 + \Phi$

$h \frac{c}{\lambda_1} = Ek_1 + \Phi$

$h \frac{c}{\lambda_2} = Ek_2 + \Phi\\\texttt{\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ }} ( - )$

$hc ( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} }) = Ek_1 - Ek_2$

$6.63 \times 10^{-34} \times 3 \times 10^8 \times ( \frac{1}{4 \times 10^{-7}} - \frac{1}{2.7 \times 10^{-7}} ) = 1.76 \times 10^{-19} - Ek_2$

$- 2.39 \times 10^{-19} = 1.76 \times 10^{-19} - Ek_2$

$Ek_2 = 1.76 \times 10^{-19} + 2.39 \times 10^{-19}$

$Ek_2 = 4.15 \times 10^{-19} \texttt{ J}$

$Ek_2 = 2.59 \texttt{ eV}$

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Learn more

• Photoelectric Effect : brainly.com/question/1408276
• Statements about the Photoelectric Effect : brainly.com/question/9260704
• Rutherford model and Photoelecric Effect : brainly.com/question/1458544

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics