Question

When electromagnetic radiation of wavelength $396\stackrel{\circ }{A}$ 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. $15.65\text{eV}$
• B. $10.48\text{eV}$
• C. $9.65\text{eV}$
• D. $16.48\text{eV}$

Answer 1

The correct option is A $15.65\text{eV}$

Given $\lambda$ = $396\stackrel{\circ }{A}=396\times {10}^{-10}$ m
From photoelectric 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,
$\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{396\times {10}^{-10}}=15.6\times 1.6\times {10}^{-19}+(KE{)}_{max}$
$\Rightarrow K.E_{max}=(0.05\times {10}^{-16}-24.96\times {10}^{-19})J$
$\Rightarrow K.E_{max}=25.04\times {10}^{-19}J=15.65eV$