Question

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

• A. $-\ge2.8 \times 10^{-9}m $
• B. $\le2.8 \times 10^{-12}m $
• C. $<2.8 \times 10^{-10}m $
• D. $<2.8 \times 10^{-9}m $

Answer 1

The Correct Option is A

: According to Einstein's photoelectric
equation, the maximum kinetic energy of the
emitted electron is
$K_{max}=\frac {hc}{?}-\phi_0$
where $? $ is the wavelength of incident light and $\phi_0 $
is the work function.
Here, $? =500 \, nm, $ he = 1240 eV nm
and $\phi_0=2.28eV$
$\therefore K_{max}= \frac {1240 \, eV \, nm}{500 \, nm}-2.28 e V $
$\, \, \, \, \, \, \, \, =2.48eV-2.28 eV=0.2eV $
The de Broglie wavelength of the emitted electron
is
$?_{min}=\frac {h}{\sqrt {2mK_{max}}} $
where h is the Planck's constant and m is the mass
of the electron.
As h$=6.6 \times 10^{-34}J \, s,\, \, m=9 \times 10^{-31} kg$
and $K_{max}=0.2eV=0.2 \times1.6 \times 10^{-19} J $
$\therefore \, \, \, ?_{min}=\frac {6.6 \times 10^{-34}J\, s}{\sqrt {2(9 \times10^{-31}kg)(0.2 \times 1.6 \times 10^{-19}J)}}$
$\, \, \, \, =\frac {6.6}{2.4} \times 10^{-9}m=2.8 \times 10^{-9} m$
So, $?\ge2.8 \times10^{-9}m$