Question

If work function of Cesium is ${\varphi }_0=1.9\text{eV}$. Find the maximum kinetic energy of the emitted electrons, if the metal is irradiated with a radiation of wavelength $700\text{nm}$.

• A. $1.3\text{eV}$
• B. $5\text{eV}$
• C. $2\text{eV}$
• D. No photoemission will occur

Answer 1

The correct option is D No photoemission will occur

Energy of incident radiation is,
$E=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{700\times {10}^{-9}}=2.828\times {10}^{-19}\text{J}\approx 1.77\text{eV}$
$\because$ Energy of incident radiation is less than the work function (${\varphi }_0$)
$\therefore$ No photoelectron emission will occur

Hence, $\left(D\right)$ is the correct answer.

Why this question? The question is a typical one concept $K.E_{max}=E-{\varphi }_0$ If $E<{\varphi }_0$, no photoemission will occur.