Question

A particle X moving with a certain velocity has a de Broglie wavelength of 1 nm. If a particle Y has a mass of 25% that of X and velocity 75% that of X, then de Broglie wavelength of Y will be:
A.3
B.5.33
C.6.88
D.48

Answer 1

Hint: We know that de-Broglie discovered the dual nature of all moving particles, that is, particle nature and wave nature. Here, we have to use the de Broglie equation to calculate the wavelength of the Y particle. The de-Broglie equation is, $\lambda = \dfrac{h}{{mv}}$. Here, h is Planck’s constant, $\lambda $ is wavelength, m is mass and v is velocity.

Complete step by step answer:
Here, two moving particles X and Y are given. The wavelength of X is given as 1 nm. The mass and velocity of particle Y is given in terms of X. So, first we have to calculate the mass of velocity of Y in terms of X.

Mass of Y=25% of Mass of X

$ \Rightarrow {\rm{Mass}}\,{\rm{of}}\,{\rm{Y}} = \dfrac{{25}}{{100}}{m_X} = 0.25{m_X}$

Velocity of Y=75% of Velocity of X
$ \Rightarrow {\rm{Velocity}}\,{\rm{of}}\,{\rm{Y}} = \dfrac{{75}}{{100}}{v_X} = 0.75{v_X}$

Now, we have to write the de Broglie equation for particle X.

${\lambda _X} = \dfrac{h}{{{m_x}{v_x}}}$

The wavelength is given as 1 nm. So, the above equation becomes,

$\dfrac{h}{{{m_x}{v_x}}} = 1$…… (1)

Now, we write the de-Broglie equation for particle Y. The mass of Y is $0.25\,{m_X}$ and velocity of Y is $0.75\,{v_x}$.

${\lambda _y} = \dfrac{h}{{{m_y}{v_y}}}$

$ \Rightarrow {\lambda _y} = \dfrac{h}{{0.25{m_x} \times 0.75{v_x}}}$

$ \Rightarrow {\lambda _y} = \dfrac{1}{{\left( {0.25 \times 0.75} \right)}} \times \dfrac{h}{{{m_x} + {v_x}}}$

From equation (1), wavelength of particle X is 1nm. So, the above equation becomes,

$ \Rightarrow {\lambda _y} = \dfrac{1}{{\left( {0.25 \times 0.75} \right)}} = 5.33\,{\rm{nm}}$

Therefore, the wavelength of particle Y is 5.33 nm.

So, the correct answer is Option B .

Note: The wave and particle nature of the moving particle is inversely proportional to each other. Mathematical expression of the above statement is,

$\lambda \alpha \dfrac{1}{p}$

Here, $\lambda $ is wavelength and p is momentum.

Always remember that de Broglie relation can be applied only to the moving microscopic particles including protons, electrons, atoms etc. It has no relevance for the moving semi micro or macroparticles.