Question

The work functions for metals A, B and C are $1.92eV,2.0eV$ and $5.0eV$ respectively. According to Einstein’s equation, the metals which will emit photoelectrons for a radiation of wavelength $4100A^{\circ }$ is/are

• A. None
• B. A only
• C. A and B only
• D. All the three metals

Answer 1

The correct option is C A and B only

Einstein's equation is
$E=\varphi +K.E$
where $E$ is the energy of incident radiation, $\varphi$ is work function of metal and $K.E$ is the kinetic energy of emitted electrons.

Thus for emission of photoelectrons, $E\ge \varphi$
In this case, energy of incident radiation
$E=\frac{hc}{\lambda }=\frac{12431}{\lambda \left(\text{in}\stackrel{\circ }{A}\right)}eV$
$=\frac{12431}{4100}=3.03eV$

$E>{\varphi }_A$ and ${\varphi }_B$ but $E<{\varphi }_C$
So, electrons will be emitted from both A and B.