Question

Ultraviolet light of wavelength $\lambda =200\text{nm}$ is used in an experiment on photoelectric effect with caesium ($\varphi =1.8\text{eV}$). Find De-Broglie wavelength of photo electron with maximum kinetic energy.

• A. $\frac{4}{7}\times {10}^{-9}m$
• B.
  $\frac{28}{5}\times {10}^{-9}m$
• C. $\frac{19}{7}\times {10}^{-9}m$
• D. $\frac{2}{7}\times {10}^{-9}m$

Answer 1

The correct option is A $\frac{4}{7}\times {10}^{-9}m$

$K{E}_{\text{max}}=hv-\varphi$

$h\nu =\frac{1242}{200}=6.21\text{eV}$
$\varphi =1.8\text{eV}$

$K{E}_{\text{max}}=6.21-1.8=4.41\text{eV}$

$\lambda =\frac{1.2}{\sqrt{4.41}}\text{nm}=\frac{1.2}{2.1}\text{nm}=\frac{4}{7}\times {10}^{-9}\text{m}$