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Light quanta with an energy 4.9 eV ejects photoelectrons from a metal with work function 4.5 eV .Then the maximum impulse transimitted to surface of metal when each electron flies out. Then de-Broglie wavelength of photoelctrons :

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

1 Answer

Answer : (b) $\;2 \times10^{-8} m$
Explanation :
Einstein photoelectric equation
$\large\frac{1}{2} mV^2=4.9-4.5=0.4 eV=0.4\times16\times10^{-19 } J$
$V=\sqrt{\large\frac{2 \varepsilon}{M}}$
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 $\;\lambda=\large\frac{h}{mV}$
$=\large\frac{6.63 \times 10^{-33}}{3.45 \times10^{-25}}=2 \times 10^{-8} m\;.$
answered Feb 26, 2014 by yamini.v

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