Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Assertion : The de - Broglie wavelength of a molecule varies inversely as the square root of temperature. Reason : The root mean square velocity of the molecule depends on the temperature.
A. If both assertion and reason are true and reason is the correct explanation of the assertion.
B. If both assertion and reason are true but reason is not the correct explanation of the assertion.
C. If the assertion is true but the reason is false.
D. If both the assertion and reason are false.

Answer
VerifiedVerified
161.4k+ views
Hint:In order to solve this question, we will first write the formula of De- Broglie wavelength and then we will check whether the assertion and reason are correct statements or not and also the correctness of reason for the given assertion.

Formula used:
The de- Broglie wavelength is given as,
$\lambda = \dfrac{h}{{mv}}$
where, h is Plank’s constant, m is the mass, and v is the velocity of the molecule.
The expression for root mean square velocity is,
${v_{r.m.s}} = \sqrt {\dfrac{{3kT}}{m}} $
Here, $m$ is the mass, $T$ is the temperature and $k$ is the boltzmann constant.

Complete step by step solution:
As we know that, the de- Broglie wavelength is given by the formula as,
$\lambda = \dfrac{h}{{mv}}$
We also know that for a molecule of a gas its velocity is directly proportional to the temperature is,
${v_{r.m.s}} = \sqrt {\dfrac{{3kT}}{m}} $
Using this, the de- Broglie wavelength will depend upon temperature as $\lambda \propto \dfrac{h}{{m\sqrt T }}$

So, our assertion: The de - Broglie wavelength of a molecule varies inversely as the square root of temperature is correct as we found $\lambda \propto \dfrac{h}{{m\sqrt T }}$
And our reason: The root mean square velocity of the molecule depends on the temperature is also correct as $v \propto \sqrt T $ and also reason is not the correct explanation of our assertion because reason does not provide how the velocity depends upon temperature

Hence, the correct answer is option B.

Note: It should be remembered that the de-Broglie wavelength is the wavelength associated with every matter in the universe and this concept shows the wave nature of matter apart from the particle nature of matter.