Question 78: Dual Nature of Matter and Radiation

Question

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave?
(Given $\mathrm{hc}=1242 \mathrm{eVnm}$ where h is the Planck's constant and c is the speed of light in vaccum.)

Accepted Answer

Let's break down the problem step by step:
In the photoelectric effect, the energy of a photon is given by:
$$ h
u = \phi + K_e $$
where:
$$\phi$$ is the work function of the metal,
$$K_e$$ is the kinetic energy of the ejected electron.
The kinetic energy of the ejected electrons is provided by the stopping potential:
$$ K_e = eV $$
Given that the stopping potential is $$ V = 2 	ext{ V} $$, we have:
$$ K_e = 2 	ext{ eV} $$
Now, we can calculate the photon energy:
$$ h
u = \phi + K_e = 2.14 	ext{ eV} + 2 	ext{ eV} = 4.14 	ext{ eV} $$
The relation between the photon's energy and its wavelength is:
$$ h
u = \frac{hc}{\lambda} $$
Rearranging for $$\lambda$$:
$$ \lambda = \frac{hc}{h
u} $$
Given that $$ hc = 1242 	ext{ eVÃÂ·nm} $$, substitute the values:
$$ \lambda = \frac{1242 	ext{ eVÃÂ·nm}}{4.14 	ext{ eV}} $$
Calculate the wavelength:
$$ \lambda \approx \frac{1242}{4.14} \approx 300 	ext{ nm} $$
Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm.
The correct answer is Option C: 300 nm.

Answer

400 nm

Answer

600 nm

Answer

300 nm

Answer

200 nm

Dual Nature of Matter and Radiation

Choose the correct answer: