Given:
$m_1 = 100 \text{ g}, \quad V_1 = V_1 \text{ m/s}$
$m_2 = 50 \text{ g}, \quad V_2 = 1.5V_1 \text{ m/s}$
To find:
$\dfrac{\lambda(m_1)}{\lambda(m_2)} = ?$
Using de-Broglie’s wavelength formula:
$\lambda = \dfrac{h}{mv}$
So, the ratio is: \[ \dfrac{\lambda_1}{\lambda_2} = \dfrac{m_2 V_2}{m_1 V_1} = \dfrac{50 \times 1.5V_1}{100V_1} = \dfrac{3}{4} \]
Final Answer: The ratio is 3 : 4.
The de Broglie wavelength ($\lambda$) of a particle is given by the formula:
$\lambda = \frac{h}{p} = \frac{h}{mv}$
where:
We are given two baseballs with masses $m_1 = 100 \text{ g}$ and $m_2 = 50 \text{ g}$. The velocity of $m_2$ is 1.5 times that of $m_1$, so $v_2 = 1.5v_1$.
We want to find the ratio of their de Broglie wavelengths: $\lambda(m_1) : \lambda(m_2)$
The de Broglie wavelength for mass $m_1$ is:
$\lambda_1 = \frac{h}{m_1v_1}$
The de Broglie wavelength for mass $m_2$ is:
$\lambda_2 = \frac{h}{m_2v_2}$
Now, let's find the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{m_1v_1}}{\frac{h}{m_2v_2}} = \frac{m_2v_2}{m_1v_1}$
Substitute the given values: $m_1 = 100$, $m_2 = 50$, and $v_2 = 1.5v_1$:
$\frac{\lambda_1}{\lambda_2} = \frac{50 \cdot 1.5v_1}{100 \cdot v_1} = \frac{75}{100} = \frac{3}{4}$
Therefore, the ratio of the de Broglie wavelengths is:
$\lambda(m_1) : \lambda(m_2) = 3 : 4$
Given:
$m_1 = 100 \, \text{g}$
$V_1 = V_1 \, \text{m/s}$
$m_2 = 50 \, \text{g}$
$V_2 = 1.5 V_1 \, \text{m/s}$
Now, using the formula for de-Broglie wavelength:
$\lambda = \frac{h}{mv}$
$\Rightarrow \frac{\lambda_1}{\lambda_2} = \frac{m_2 V_2}{m_1 V_1} = \frac{50 \times 1.5 V_1}{100 V_1} = \frac{3}{4}$
So, the correct option is (B) : 3 : 4
If \( \lambda \) and \( K \) are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ‘v’ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.