Question

Light of wavelength $500nm$ is incident on a metal with a work function of $2.28eV$. The de borglie wavelength of the emitted electron is

• A. $\le 2.8\times {10}^{-12}m$
• B. $<2.8\times {10}^{-10}m$
• C. $<2.8\times {10}^{-9}m$
• D. $\ge 2.8\times {10}^{-9}m$

Answer 1

The correct option is D $\ge 2.8\times {10}^{-9}m$

From photoelectric effect, $h\nu =W_0+eV$

or $\frac{hc}{\lambda }=W_0+eV$

or $\frac{(6.6\times {10}^{-34})\times (3\times {10}^8)}{500\times {10}^{-9}}$
$=2.28\times 1.6\times {10}^{-19}+eV$
or $eV=0.31\times {10}^{-19}$ or $V=0.2$

Here $V$ should be $V\le 0.2$

The de Broglie wavelength of electron , $\lambda =\frac{12.27\times {10}^{-10}}{V}$

So, $\lambda \ge 2.8\times {10}^{-9}m$