Question

A metal is irradiated with light of wavelength $600nm$. Given that the work function of the metal is $1.0eV$, 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. $6.6\times {10}^{-13}m$

Answer 1

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

$\lambda =600nm=600\times {10}^{-9}m$

$w=1.0ev=1.0\times {10}^{-19}\times 1.6joules={hv}_0$

$hv={hv}_0+k.E$

$K.E=\frac{p^2}{2m}=\frac{h^2}{{2m\lambda }^2}$

We know that debroglies wavelength$\left(\lambda \right)$

$\left(\lambda \right)=\frac{h}{p}$

$E=hv=\frac{hc}{\lambda }=\frac{\left({6.625\times 10}^{-34}\right)\left({3\times 10}^8\right)}{\left({600\times 10}^{-9}\right)}$

$=3.313\times {10}^{-9}joules$

$K.E=hv-{hv}_0$

$K.E=E-W$

$=(3.313-1.6){10}^{-19}$

$\frac{h^2}{{2m\lambda }^2}=1.713\times {10}^{-19}joules$

${\lambda }^2=\frac{{(6.626\times {10}^{-34})}^2}{2(9.1\times {10}^{-31})\left({1.713\times 10}^{-19}\right)}$

$\lambda =1.28\times {10}^{-9}m$

Debroglies wavelength $\approx 1.3\times {10}^{-9}m$