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Physics Question on Dual nature of matter

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

−≥2.8×10−9m-\ge2.8 \times 10^{-9}m

B

≤2.8×10−12m\le2.8 \times 10^{-12}m

C

<2.8×10−10m<2.8 \times 10^{-10}m

D

<2.8×10−9m<2.8 \times 10^{-9}m

Answer

−≥2.8×10−9m-\ge2.8 \times 10^{-9}m

Explanation

Solution

: According to Einstein's photoelectric
equation, the maximum kinetic energy of the
emitted electron is
Kmax=hc?−ϕ0K_{max}=\frac {hc}{?}-\phi_0
where ?? is the wavelength of incident light and ϕ0\phi_0
is the work function.
Here, ?=500 nm,? =500 \, nm, he = 1240 eV nm
and ϕ0=2.28eV\phi_0=2.28eV
∴Kmax=1240 eV nm500 nm−2.28eV\therefore K_{max}= \frac {1240 \, eV \, nm}{500 \, nm}-2.28 e V
        =2.48eV−2.28eV=0.2eV\, \, \, \, \, \, \, \, =2.48eV-2.28 eV=0.2eV
The de Broglie wavelength of the emitted electron
is
?min=h2mKmax?_{min}=\frac {h}{\sqrt {2mK_{max}}}
where h is the Planck's constant and m is the mass
of the electron.
As h=6.6×10−34J s,  m=9×10−31kg=6.6 \times 10^{-34}J \, s,\, \, m=9 \times 10^{-31} kg
and Kmax=0.2eV=0.2×1.6×10−19JK_{max}=0.2eV=0.2 \times1.6 \times 10^{-19} J
∴   ?min=6.6×10−34J s2(9×10−31kg)(0.2×1.6×10−19J)\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)}}
    =6.62.4×10−9m=2.8×10−9m\, \, \, \, =\frac {6.6}{2.4} \times 10^{-9}m=2.8 \times 10^{-9} m
So, ?≥2.8×10−9m?\ge2.8 \times10^{-9}m