Question

When ultraviolet light with a wavelength of 254 nm falls on a clean copper surface, the stopping potential necessary to stop emission of photoelectrons is 0.181 v. part a what is the photoelectric threshold wavelength for this copper surface?

Answer 1

The ultraviolet wavelength λ = 254nm the stopping potential of copper surface,
V₀ = 0.18W
Now the relation  between threshold wavelength and stopping potential is
eV₀ = λc (1/λ - 1/λ₀)
1/λ - 1/λ₀ = eV₀/λc
1/λ₀ = 1/λ eV/λc
= 1/ 254 × 10⁻⁹m - 1.6 10⁻¹⁹C)(0.181v)/6.625 × 10⁻³⁴ Js (3×10⁸m/s
0.00393 × 10⁹ - 0.01457×10⁷
= 0.378 ×10⁷
λ₀ = 264 × 10⁻⁹m or
264nm 

Answer 2

The photoelectric threshold wavelength for this copper surface is 264 nm

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 )

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 !

Given:

λ = 254 nm = 2,54 × 10⁻⁷ m

V = 0.181 Volt

c = 3 × 10⁸ m/s

h = 6.63 × 10⁻³⁴ Js

q = 1.6 × 10⁻¹⁹ C

Unknown:

λ₀ = ?

Solution:

$E = qV + \Phi$

$h f = qV + h f_o$

$h \frac{c}{\lambda} = qV + h \frac{c}{\lambda_o}$

$6.63 \times 10^{-34} \times \frac{3 \times 10^8}{2.54 \times 10^{-7}} = 1.6 \times 10^{-19}(0.181) + 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda_o}$

$7.83 \times 10^{-19} = 2.896 \times 10^{-20} + \frac{1.989 \times 10^{-25}}{\lambda_o}$

$\frac{1.989 \times 10^{-25}}{\lambda_o} = 7.83 \times 10^{-19} - 2.896 \times 10^{-20}$

$\lambda_o \approx 2.64 \times 10^{-7} ~ m$

$\large {\boxed{\lambda_o \approx 264 ~ nm} }$

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

Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics

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