Question

A light of wavelength 400 nm is incident on a metal with a work function 2.28 eV. The de-broglie wavelength of the emitted electron is:

• A. 17140 $\stackrel{\circ }{A}$
• B. 18140 $\stackrel{\circ }{A}$
• C. 16140 $\stackrel{\circ }{A}$
• D. 15140 $\stackrel{\circ }{A}$

Answer 1

The correct option is D 15140 $\stackrel{\circ }{A}$

Energy of Incident radiation according to planck's quantum theory E = $\frac{hc}{\lambda }$
where $\lambda$ = wavelength of incident radiation
E = $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}$
= $4.96\times {10}^{-19}J$
According to Photo electric effect:
$K.E_{max}=E-W$
where E = Energy of incident radiation
W = work function of the metal
$W=2.28eV=2.28\times 1.6\times {10}^{-19}$
= $3.65\times {10}^{-19}J$
So, $K.E_{max}=4.96\times {10}^{-19}-3.65\times {10}^{-19}$ = $1.31\times {10}^{-19}$ J
= $\frac{1.31\times {10}^{-19}}{1.6\times {10}^{-19}}$
= 0.819 eV
Using relation between kinetic energy and de-broglie wavelength
E (eV) = $\frac{12400}{\lambda }$ $\stackrel{\circ }{A}$
So, $\lambda$ = $\frac{12400}{0.819}\stackrel{\circ }{A}$
$\lambda$ = 15140.42 $\stackrel{\circ }{A}$ =$15140\stackrel{\circ }{A}\left(approx\right)$