Question

What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?

Answer 1

Answer: $6.268(10)^{-16}J$

The kinetic energy of an electron $K_{e}$ is given by the following equation:

$K_{e}=\frac{(p_{e})^{2} }{2m_{e}}$ (1)

Where:

$K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}$

$p_{e}$ is the momentum of the electron

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

From (1) we can find $p_{e}$:

$p_{e}=\sqrt{2K_{e}m_{e}}$ (2)

$p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}$  

$p_{e}=2.091(10)^{-24}\frac{kgm}{s}$ (3)

Now, in order to find the wavelength of the electron $\lambda_{e}$ with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

$\lambda_{e}=\frac{h}{p_{e}}$ (4)  

Where:  

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant  

So, we will use the value of $p_{e}$ found in (3) for equation (4):

$\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}$  

$\lambda_{e}=3.168(10)^{-10}m$ (5)  

We are told the wavelength of the photon $\lambda_{p}$ is the same as the wavelength of the electron:

$\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m$ (6)  

Therefore we will use this wavelength to find the energy of the photon $E_{p}$ using the following equation:

$E_{p}=\frac{hc}{lambda_{p}}$ (7)  

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

$E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}$  

Finally:

$E_{p}=6.268(10)^{-16}J$

Answer 2

The energy of a photon is about 6.3 × 10⁻¹⁶ Joule

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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:

kinetic energy of an electron = Ek = 15 eV = 2.4 × 10⁻¹⁸ Joule

Asked:

energy of photon = E = ?

Solution:

Firstly , we will use the formula of kinetic energy:

$Ek = \frac{1}{2}mv^2$

$v^2 = \frac{2Ek}{m}$

$v = \sqrt{ \frac{2Ek}{m}}$

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Next , we will use the formula of The Broglie's Wavelength:

$\lambda = \frac{h}{mv}$

$\lambda = \frac{h}{m\sqrt{2Ek/m}}$

$\lambda = \frac{h}{\sqrt{2mEk}}$

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$E = h f$

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

$E = h \frac{c}{\frac{h}{\sqrt{2mEk}}}$

$E = c\sqrt{2mEk}$

$E = 3 \times 10^8 (\sqrt{2(9.11 \times 10^{-31} \times 2.4 \times 10^{-18}}$

$E \approx 6.3 \times 10^{-16} \texttt{ Joule}$

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Learn more

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

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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