Dual Nature Of Radiation And Matter Ques 73
73. Photoelectric work function of a metal is $1 $ $eV$. Light of wavelength $\lambda=3000$ $ \AA$ falls on it. The photo electrons come out with velocity
[1991]
(a) $10$ $ metres / sec$
(b) $10^{2}$ $metres / sec$
(c) $10^{4}$ $ metres / sec$
(d) $10^{6}$ $ metres / sec$
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Answer:
Correct Answer: 73.(d)
Solution:
- (d) $h v=W+\frac{1}{2}$ $ m v^{2}$ or
$\frac{h c}{\lambda}=W+\frac{1}{2} $ $m v^{2}$
Here $\lambda=3000 $ $\AA=3000 \times 10^{-10} $ $m$
and $W=1 eV=1.6 \times 10^{-19}$ joule
$ \begin{aligned} & \therefore \frac{(6.6 \times 10^{-34})(3 \times 10^{8})}{3000 \times 10^{-10}} \\ & =(1.6 \times 10^{-19})+\frac{1}{2} \times(9.1 \times 10^{-31}) v^{2} \end{aligned} $
Solving we get, $v \cong 10^{6} $ $m / s$
BETA
