Question

An axial 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:

• A. $6.6\times {10}^{-7}m$
• B. $1.3\times {10}^{-9}m$
• C. $8.9\times {10}^{-11}m$
• D. $6.6\times {10}^{-13}m$

Answer 1

The correct option is B $1.3\times {10}^{-9}m$

Energy of the Photon$=\frac{1240}{660}=1.88eV$

Kinetic energy of photoelectron emitted $=(1.88-1.0)=0.88eV$

$\lambda =\frac{h}{\sqrt{2.K.E.m}}=\frac{6.62\times {10}^{-34}}{\sqrt{2\times 0.88\times 1.6\times {10}^{-19}\times 9.1\times {10}^{-31}}}=1.3\times {10}^{-9}m$