JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 4)
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)
(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)
0.44 $$\mathop A\limits^o $$
0.34 $$\mathop A\limits^o $$
0.20 $$\mathop A\limits^o $$
0.24 $$\mathop A\limits^o $$
Explanation
We know, the de-Broglie
wavelength
$$\lambda $$ = $${h \over {m{v_{rms}}}}$$
also Vrms = $$\sqrt {{{3kT} \over m}} $$
$$ \therefore $$ $$\lambda $$ = $${h \over {\sqrt {3mkT} }}$$
= $${{6.63 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 4.6 \times {{10}^{ - 26}} \times 1.38 \times {{10}^{ - 23}} \times 400} }}$$
= 2.4 $$ \times $$ 10-11 m
= 0.24 $$\mathop A\limits^o $$
$$\lambda $$ = $${h \over {m{v_{rms}}}}$$
also Vrms = $$\sqrt {{{3kT} \over m}} $$
$$ \therefore $$ $$\lambda $$ = $${h \over {\sqrt {3mkT} }}$$
= $${{6.63 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 4.6 \times {{10}^{ - 26}} \times 1.38 \times {{10}^{ - 23}} \times 400} }}$$
= 2.4 $$ \times $$ 10-11 m
= 0.24 $$\mathop A\limits^o $$
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