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.)

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