When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _______$\mathop A\limits^o$. (Round off to the Nearest Integer).
[ Use : $\sqrt 3$ = 1.73, h = 6.63 $\times$ 10$-$34 Js
me = 9.1 $\times$ 10$-$31 kg; c = 3.0 $\times$ 108 ms$-$1; 1eV = 1.6 $\times$ 10$-$19 J]
Answer (integer)
9
Solution
$\lambda$ = 248 $\times$ 10<sup>$-$9</sup> m<br><br>w<sub>0</sub> = 3 $\times$ 1.6 $\times$ 10<sup>$-$19</sup> J<br><br>E = w<sub>0</sub> + K.E.<br><br>${{hc} \over \lambda } = {W_0} + K.E.$<br><br>$$K.E. = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {248 \times {{10}^{ - 19}}}} - 3 \times 1.6 \times {10^{ - 19}}$$<br><br>= 3.2 $\times$ 10<sup>$-$19</sup> J<br><br>p = $\sqrt {2mK.E.}$<br><br>p = $\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 3.2 \times {{10}^{ - 19}}}$<br><br>p = 7.63 $\times$ 10<sup>$-$25</sup><br><br>$\therefore$ $$\lambda = {h \over p} = {{6.626 \times {{10}^{ - 34}}} \over {7.63 \times {{10}^{ - 25}}}}$$<br><br>$\Rightarrow$ $\lambda$ = 8.7 $\times$ 10<sup>$-$10</sup> = 8.7 $\mathop A\limits^o$ $\approx$ 9 $\mathop A\limits^o$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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