Question

When electromagnetic radiation of wavelength $500\stackrel{\circ }{A}$ is incident on a metal sheet that has a work function of 15.6 eV, find the maximum kinetic energy of the photoelectrons emitted.
Take $h=6.6\times {10}^{-34}Js$

• A. 7.18 eV
• B. 11.1 eV
• C. 9.15 eV
• D. 5.5 eV

Answer 1

The correct option is C 9.15 eV

Given $\lambda$ = $500\stackrel{\circ }{A}=500\times {10}^{-10}$ m
From photo-electric effect,
$h\nu =W+(KE{)}_{max}$
$\Rightarrow$$\frac{hc}{\lambda }=W+(K.E{)}_{max}$
where $\lambda$ = wavelength of incident radiation
W = work function of the metal sheet
$(KE{)}_{max}$ = Kinetic energy of the emitted photoelectron.
So,
$\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-10}\times 1.6\times {10}^{-19}}$ = $15.6+(KE{)}_{max}$
On solving, $(KE{)}_{max}$ = $9.15eV$