Question

The photoelectric threshold wavelength of tungsten is $230nm$. The energy of electrons ejected from its surface by ultraviolet light of wavelength $180nm$ is

• A. $0.15eV$
• B. $1.5eV$
• C. $15eV$
• D. $15keV$

Answer 1

The correct option is B $1.5eV$

Phototelectric law,

$KE=\frac{hc}{\lambda }-W=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{\circ }}where,W=\frac{hc}{{\lambda }_{\circ }}and{\lambda }_{\circ }isthresholdwavelength\therefore KE=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_{\circ }})=\frac{6.6\times {10}^{-34}\times 3\times {10}^8\times {10}^9(\frac{1}{180}-\frac{1}{230})}{1.6\times {10}^{-19}}=1.5eV$