Question

If the de-Broglie wavelength of a particle of mass m is $100$ times its velocity than its value in terms of its mass (m) and Plank’s constant (h) is:
A. $\dfrac{1}{{10}}\sqrt {\dfrac{{\text{m}}}{{\text{h}}}} $
B. $10\sqrt {\dfrac{{\text{h}}}{{\text{m}}}} $
C.$\dfrac{1}{{10}}\sqrt {\dfrac{{\text{h}}}{{\text{m}}}} $
D. $10\sqrt {\dfrac{{\text{m}}}{{\text{h}}}} $

Answer 1

Hint: We can use the de-Broglie equation which represents the relation between wavelength, mass of the particle, velocity of the particle and Plank’s constant. Firstly write the equation of the planck and then put the given value to ultimately find the required relation.

Complete step by step solution:
The waves have a dual nature. In $1924$ De-Broglie told that the matter also has dual nature. The matter that has linear momentum also has a wave associated with it. The De-Broglie equation gives the relation between wave nature and particle nature. From the known wavelength its mass can be determined and from the given mass the wavelength can be determined.
De-Broglie equation is represented as follows:
$\lambda = \,\dfrac{{\text{h}}}{{{\text{mv}}}}$
$\lambda $ is the de-Broglie wavelength of the particle.
${\text{h}}$is the Planck's constant.
${\text{m}}$is the mass of the particle.
${\text{v}}$ is the velocity of the particle.
It is given that wavelength is $100$times its velocity so,
$\lambda = 100\,{\text{v}}\,$
On rearranging for ${\text{v}}$we get,
${\text{v}}\, = \,\dfrac{\lambda }{{100}}\,$
On substituting the value of ${\text{v}}$in De-Broglie equation we get,
$\Rightarrow \lambda = \,\dfrac{{\text{h}}}{{{\text{m}}\dfrac{\lambda }{{100}}\,}}$
$\Rightarrow \lambda = \,\dfrac{{100\,{\text{h}}}}{{{\text{m}}\lambda \,}}$
$\Rightarrow {\lambda ^2} = \,100\dfrac{{\,{\text{h}}}}{{{\text{m}}\,}}$
$\Rightarrow \lambda \, = 10\sqrt {\dfrac{{\text{h}}}{{{\text{m}}\,}}} $
So, the wavelength in terms of its mass (m) and Plank’s constant (h) is $10\sqrt {\dfrac{{\text{h}}}{{{\text{m}}\,}}} $.

Therefore, option (B) $10\sqrt {\dfrac{{\text{h}}}{{{\text{m}}\,}}} $, is correct.

Note: The product of mass and velocity is given as momentum.so, the De-Broglie equation also written as $\lambda = \,\dfrac{{\text{h}}}{{\text{p}}}$ where, ${\text{p}}$ is the momentum. The de-Broglie wavelength of the particle is inversely proportional to its momentum. The unit of wavelength is meter. The unit of velocity is meter/second. The value of Planck's constant is $6.6 \times {10^{ - 34}}{\text{J}}\,{\text{s}}$.