KVPY SX SB Exam 2012 Question Paper

Question : 105 of 120

Marks: +1, -0

A metal is irradiated with light of wavelength 660 nm. Given that the work function of the metal is 1.0 eV, the de Broglie wavelength of the ejected electron is close to

$6.6 ×10^{-7} m$
$8.9 ×10^{-11} m$
$1.3×10^{-9}m$
$6.6×10^{-13} m$

Validate

Solution:

E = ϕ + K.E.

∵\E=\hC/λ

={6.6×10^{-34}×3×10^8}/{660×10^{-9}}

=3×10^{-19}\J

ϕ=1\ev=1.6×10^{-19}\J

\K.\E.=3×10^{-19}-1.6×10^{-19}

=1.4×10^{-19}\J

for wave length of emitted electron

\λ=h/√{2mKE}

={6.6×10^{-34}}/√{2×9.1×10^{-31}×1.4×10^{-19}}

=1.32×10^{-9}\meter

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