Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength
of nitrogen molecule is close to :
(Given : nitrogen molecule weight : 4.64 $\times$ 10–26 kg,
Boltzman
constant: 1.38 $\times$ 10–23 J/K,
Planck constant : 6.63 $\times$ 10–34 J.s)
Solution
We know, the de-Broglie
wavelength<br><br>
$\lambda$ = ${h \over {m{v_{rms}}}}$
<br><br>also V<sub>rms</sub> = $\sqrt {{{3kT} \over m}}$
<br><br>$\therefore$ $\lambda$ = ${h \over {\sqrt {3mkT} }}$
<br><br>= $${{6.63 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 4.6 \times {{10}^{ - 26}} \times 1.38 \times {{10}^{ - 23}} \times 400} }}$$
<br><br>= 2.4 $\times$ 10<sup>-11</sup> m
<br><br>= 0.24 $\mathop A\limits^o$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
This question is part of PrepWiser's free JEE Main question bank. 145 more solved questions on Dual Nature of Matter and Radiation are available — start with the harder ones if your accuracy is >70%.