Question

A metal is irradiated with a 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

• A. $6.6\times 10^{- 7}m$
• B. $8.9\times 10^{- 11}m$
• C. $1.3\times 10^{- 9}m$
• D. $6.6\times 10^{- 33}m$

Answer 1

Answer: (C)

We know the kinetic energy of the electron is given by:
K.E. $=E_p-\phi K . E .=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} J -1.6 \times 10^{-19} JK .$
$ E. =1.4 \times 10^{-19} J$
And the de-broglie wavelength is given by:
$\lambda=\frac{ h }{\sqrt{2 mK . E .}} \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.4 \times 10^{-19}}}=1.3 \times 10^{-9} m$