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Photoelectrons are liberated by ultraviolet light of wavelength $3000A^{\large\circ}$ from a metallic surface for which the photoelectric threshold is 4000$A^{\large\circ}$.Calculate De-Broglie wavelength of electrons emitted with maximum kinetic energy.

$\begin{array}{1 1}(a)\;1.2\times 10^{-7}m&(b)\;1.2\times 10^{-9}m\\(c)\;2.3\times 10^{-9}m&(d)\;1\times 10^{-3}m\end{array}$

1 Answer

KE =Quantum energy-Threshold energy
$\Rightarrow \large\frac{6.626\times 10^{-34}\times 3\times 10^8}{3000\times 10^{-10}}-\frac{6.626\times 10^{-34}\times 3\times 10^8}{4000\times 10^{-10}}$
$\Rightarrow 6.626\times 10^{-19}-4.9695\times 10^{-19}$
$\Rightarrow 1.6565\times 10^{-19}$J
$\large\frac{1}{2}$$mv^2=1.6565\times 10^{-19}$
$m^2v^2=2\times 1.6565\times 10^{-19}\times 9.1\times 10^{-31}$
$mv=5.49\times 10^{-25}$
$\lambda=\large\frac{h}{mv}$
$\Rightarrow \large\frac{6.626\times 10^{-34}}{5.49\times 10^{-25}}$
$\Rightarrow 1.2\times 10^{-9}m$
Hence (b) is the correct answer.
answered Jan 17, 2014 by sreemathi.v
 

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