Question

The wavelength associated with a golf ball weighing 200g and moving at a speed of 5m/h is of the order:
a)${{10}^{-10}}m$
b)${{10}^{-20}}m$
c)${{10}^{-30}}m$
d)${{10}^{-40}}m$

Answer 1

Hint: The answer to this question is based on the basic concept of chemistry that deals with the De - Broglie wavelength and is given by the formula$\lambda =\dfrac{h}{mv}$ and this gives the required answer.

Complete answer:
The concepts of the chemistry from the lower classes that deal with the topic of calculation of wavelength according to several formulas are familiar to us.
Let us now deal with the concept that includes De - Broglie wavelength which is well known to us.
- De – Broglie wavelength is associated with the wavelength of the wave like particles and is usually represented by the symbol$\lambda $.
- Here the De – Broglie derived equation that relates wavelength of the particles to that of momentum ‘p’ and hence it is inversely proportional to the momentum and can be written as,$\lambda \alpha \dfrac{1}{p}$ and thus, the constant given is the Planck’s constant denoted by ‘h’ which has the value of $6.626\times {{10}^{-34}}Js$
Now, since momentum ‘p’ is the product of mass and velocity of the particle, therefore the De – Broglie wavelength associated with that particle is given by, $\lambda =\dfrac{h}{mv}$
By substituting the values, we get
\[\lambda =\dfrac{6.626\times {{10}^{-34}}Js}{200\times {{10}^{-3}}g\times \left( \dfrac{5}{60\times 60} \right)}\] [since 5m/h = 5/(3600) m/s]
Therefore, by solving the above equation, we have
\[\lambda =2.4\times {{10}^{-30}}m\]

Thus, the correct answer is option c) wavelength associated with golf ball is of the order ${{10}^{-30}}m$

Note: Note that de – Broglie wavelength is also denoted by the symbol ${{\lambda }_{dB}}$ and not just by the symbol $\lambda $ and when questions are given on the basis of this symbol directly then it is not be confused.