$(a)\;2 \times10^{-6} m\qquad(b)\;2 \times10^{-8} m\qquad(c)\;2 \times10^{-9} m\qquad(d)\;2 \times10^{-10} m$

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Answer : (b) $\;2 \times10^{-8} m$

Explanation :

Einstein photoelectric equation

$\large\frac{1}{2}\;mV^2=hf-W$

$\large\frac{1}{2} mV^2=4.9-4.5=0.4 eV=0.4\times16\times10^{-19 } J$

$V=\sqrt{\large\frac{2 \varepsilon}{M}}$

The momentum charge that is equal to impulse

$l=mV=m\;\sqrt{\large\frac{2 \varepsilon}{m}}\sqrt{2m \varepsilon}$

$I=\sqrt{2 \times 9.1\times10^{-31}\times0.4\times1.6\times10^{-19}}$

$I=mV=3.45 \times10^{-25}\;Kgm/second$

The de-Broglie wavelength $\;\lambda=\large\frac{h}{mV}$

$=\large\frac{6.63 \times 10^{-33}}{3.45 \times10^{-25}}=2 \times 10^{-8} m\;.$

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